Introduction
There is a deeper thing going on in this question. There always is! For that, however, we must look at what can be generalized in the proof of this proposition. At points of time, I'll put words into quote boxes, that we will look at as KEY aspects of this problem.
Let me first provide a solution, in which I'll emphasise the key points of the general theory of uniform convergence of iterative mappings in metric spaces, which is what this topic falls under (more general forms of convergence can be considered, for example for holomorphic functions we could have UCC type convergence, but this is basically the point).
Proof with key points highlighted
First, let's imagine that convergence holds and that the limit is continuous, and think about where this sequence should go. Uniform convergence implies pointwise convergence. In particular, fixing an $x \in (-\infty,\infty)$ we know that $\lim_{n \to \infty} f_n(x) = y$ for some $y$. In particular, by continuity of $f$ at $y$, $$
f(y) = f(\lim_{n \to \infty} f_n(x)) = \lim_{n \to \infty} f(f_n(x)) = \lim_{n \to \infty} f_{n+1}(x) = y
$$
Uniform convergence implies pointwise convergence
Continuity
and therefore $f(y) = y$.
The convergence of iterative mappings are brought back to fixed points of the iterative map.
So, when is $4^{-x}=x$? The LHS is always positive so $x>0$ must occur, and if $x\geq 1$ then $4^{-x}<1$ (Note : at this step we used the monotonicity of $f$) so there isn't any solution bigger than $1$. Thus, any solution lies in $(0,1)$.
monotonicity
We let $g(x) = x -4^{-x}$. If $f$ has two real roots, then $g(x)$ has two distinct zeros and therefore $g'(x)$ has a root in $(0,1)$, however $g'(x) = 1 + 4^{-x}\ln 4$ is positive in this range. It follows that $g(x)$ has at most one root in this range. We already know that $g(\frac 12) = 0$ so it follows that $\frac 12$ is the only real fixed point of $f$.
It follows that the constant function $\frac 12$ is the only candidate function that the iteration of $f$ can converge to.
Why must the sequence $f_n(x)$ go anywhere? Let's look quickly at how restrictive $f$ is, as an iterated function. Note that $x \in \mathbb R$, and $f_1(x)$ can be anywhere between $0$ and $\infty$.
But now, let's look at $f_2(x) = f(f(x)) = 4^{-4^{-x}}$. Since $4^{-x}>0$, $-4^{-x} <0$ so $f(f(x))< 1$ for all $x$, and of course $f(f(x)) >0$ for all $x$. We've realized that $f_2(x) \in (0,1)$ for ALL $x$ Furthermore, iterating $f$ again and again shows that $f_m(x) \in (0,1)$ for all $x$ and all $m \geq 2$. This property of a function generalizes to a much, much weaker assumption on $f$ : the reducing ranges of the iterates of $f$, if well controlled, will contribute to uniform convergence. Why is that?
It's basically because if $f$ is well-behaved on one of these iterated ranges, then it follows that its behaviour there will be sufficient to prove the result!
Iterates converge or not based on how restrictive the successive ranges of $f$ are. One restricted range where $f$ behaves supremely well is an excellent find.
But the point is, what do we mean by well-behaved? What if the interval $(0,1)$ is still not good enough for our problem?
Well, the answer is, in the spirit of the Banach fixed point theorem, that $f(x)-f(y)$ must depend on $x-y$ in a "contractive" fashion. Why don't we try seeing the derivative of $f$?
So $f'$ is $4^{-x}\ln 4$. Note that $f'$ is a positive decreasing continuous function, and for contractivity, we want to understand where $f'<1$. We can easily do so : $f'=1$ gives $x = \frac{\ln \ln 4}{\ln 4}\approx 0.235$, hence $f$ is contractive in any interval of the form $(a, \infty)$ for $a>0.25$.
Contractivity, and the derivative for proving it.
So $(0,1)$ is clearly not a good enough range if we wanted to make sure that $f$ is contractive on it. How about $f_3(x)$ then? Well, $4^{-1} = \frac 14 > 0.235$, and $4^{-0} = 1$, so it follows that $f_3(x) \in (0.235,1)$ for all $x$, where $f$ IS contractive.
What we've done, is find an index $N$ at which the values that $f_N$ can take are values at which $f$ is very well behaved.
Now, let's begin the proof.
Indeed, we always start by proving that the sequence $f_n$ is uniformly Cauchy in such a case, and it turns out that uniformly Cauchy sequences (in this case) are uniformly convergent to a unique function.
Uniformly Cauchy functions with images in a complete metric space, will converge uniformly to a unique function.
Uniformly Cauchy means : for all $\epsilon>0$ there exists $N$ such that $n,m>N$ implies $|f_n(x)-f_m(x)|<\epsilon$. To prove this, we want $N>3$, so that for $n,m > N$ , $f_n(x),f_m(x)$ will definitely lie in $(0.236,1)$ for all $x$ and we can work in this interval. Without loss of generality, let $n>m$. We first write
$$
f_{n}(x) - f_m(x) = f_{n-3}(f_3(x)) - f_{m-3}(f_3(x))
$$
To emphasise the fact that $f_3$ is well controlled. Now, we use the fundamental triangle inequality for iterated functions , which is just the telescoped sum with the triangle inequality :
$$
f_{n-3}(f_3(x)) - f_{m-3}(f_3(x)) \leq |f_{n-3}(f_{3}(x)) - f_{n-4}(f_3(x))| + |f_{n-4}(f_4(x)) - f_{n-5}(f_3(x))| + \ldots + |f_{m-2}(f_3(x)) - f_{m-3}(f_3(x))|
$$
The fundamental telescoping identity for iterative sequences.
Now, let's use the contractivity of $f$ on the range of $f_3$. There is a constant $C<1$ such that $|f(x)-f(y)|< C|x-y|$ for all $x,y$ in the range of $f_3$. As in the Banach FPT, the advantage of iteration is that because the ranges of $f_n$ are also contained in the ranges of $f_3$, we get by repeating the contractivity again and again :
$$
|f_{n-3}(f_3(x)) - f_{m-3}(f_3(x))| \leq |f_{n-3}(f_{3}(x)) - f_{n-4}(f_3(x))| + |f_{n-4}(f_3(x)) - f_{n-5}(f_3(x))| + \ldots + |f_{m-2}(f_3(x)) - f_{m-3}(f_3(x))| \\ < C^{n-4}|f_4(x) - f_3(x)| + ... + C^{m-3}|f_4(x) - f_3(x)| \\ = (C^{n-4} + ... + C^{m-3})|f_4(x) - f_3(x)| \\ = \frac{C^{n-3} - C^{m-3}}{C-1}|f_4(x) - f_3(x)| = C^{m-3}\frac{C^{n-m}-1}{C-1} |f_4(x) - f_3(x)| \leq \frac{C^{m-3}}{C-1} |f_4(x) - f_3(x)| \leq \frac{3C^{m-3}}{4(C-1)}
$$
because $f_4(x),f_3(x)$ both lie in $(\frac 14,1)$ so they both can differ by at most $\frac 34$. Thus, we've proved that :
$$
|f_n(x) - f_m(x)| \leq \frac{3C^{m-3}}{4(C-1)}
$$
where we've got a quantity that depends ONLY on $m$ : now of course, for $\epsilon>0$ take $N$ such that the RHS is smaller than $\epsilon$ for $N$ in place of $m$. (Note that RHS $\to 0$ as $m \to \infty$, by our choice of $C$).
Iteration of contractivity and summability of the geometric sequence.
Finally, uniform Cauchyness and the existence of a unique candidate limit tells us that $f_n(x) \to \frac 12$ uniformly on $(-\infty,\infty)$. $\blacksquare$
A more general situation
I've highlighted everything that one can generalize, frankly. Let me now speak about a more general situation.
Suppose that you have a complete metric space $(M,d)$ and a function $f : M \to M$. We want conditions on when $f^{(n)}$ converges. For this, by our logic in the beginning of our first problem, we know that
$f$ must have a unique fixed point.
Furthermore, we want $|f(x)-f(y)| < g(|x-y|)$ where $g$ has some properties that allow the iterates to decay quickly.
Now , if we stop at just pointwise convergence, then we've got a family of results that are stronger than Banach. Focusing on just uniform convergence, the "theorem" that we used above generalizes to the following ultra-general result :
Let $(M,d)$ be a complete metric space , and $f : M \to M$ be a continuous function , such that for some $N \in \mathbb N$ , $f$ is contractive on the image of $f^N$, and furthermore this image is a bounded set. Then, $f$ has a unique fixed point $x^*$, and furthermore the sequence of functions $f^{n}$ converge uniformly to the function $g(x) = x^*$ for all $x \in M$.
We used this theorem, with $(M,d)$ being the real numbers, $f(x) = 4^{-x}$, and $N = 3$ above. Let's prove it.
Proof : Note that $f$ has a unique fixed point in the image of $f^N$ (denoted $im(f^N)$), since the restriction of $f$ to $im(f^N)$ given by $f|_{im(f^N)}$ has image in $im(f^N)$ (why?) and is a contraction there. However, $f$ doesn't have any other fixed point : note that if $f(y) = y$ then $f^{N}(y) = y$ so $y \in im(f^N)$, where $y$ must equal to that unique fixed point in this set. So we know that if $f^{(n)}$ converges somewhere uniformly, then it must converge to the constant function given by the fixed point $x^*$.
Let the contractive constant be $C<1$. A computation similar to what we did (and everything goes through!) shows that:
$$
|f_n(x) - f_m(x)| \leq \frac{C^{m-3}}{C-1} \max_{x,y \in S} |x-y|
$$
and therefore, for $\epsilon>0$ choose the $M$ (can't use $N$!) such that RHS$<\epsilon$, and we are done! $\blacksquare$
Comments on each of the key points
Let me now speak briefly on how each of the block-quoted terms generalize (or if they do at all!)
Uniform convergence implies pointwise convergence : Always the case , note that this is related to the point of "uniform convergence of iterates come back to fixed points of the original function".
Continuity : This, in fact, often becomes a consequence of contraction-type assumptions. Indeed, a typical contraction-type assumption is that $|f(x)-f(y)| \leq g(|x-y|)$ where $g$ is a function such that $g(x) \to 0$ as $x \to 0$. If this is true, then $f$ is continuous on its domain, all the time. So it's pretty much a fixture in fixed point results.
Monotonicity : Nowhere did I use monotonicity in the theorem above, but I'm going to state and prove a result later on , that assumes monotonicity without contractivity, and shows its strength. Let's see if we can use it to prove our result as well!
Iterates converge or not based on how restrictive the successive ranges of $f$ are. One restricted range where $f$ behaves supremely well is an excellent find : A principle that never changes, and is used heavily in these kinds of results.
Contractivity : This can be weakened and well-generalized. The most generalized weakening is to pseudo-contractive mappings i.e. such that $|f(x)-f(y)| \leq |x-y|$ for all $x,y$, but then conditions need to be very strong to ensure uniform convergence. Note that it can be done in such a way that the derivative isn't even involved, since $f$ need not be assumed differentiable.
Uniformly Cauchy functions with images in a complete metric space, will converge uniformly to a unique function : Not generalizing.
The fundamental telescoping identity for iterative sequences : The identity is an ever-present, always used to prove these kinds of results.
Iteration of contractivity and summability of the geometric sequence : This is important for our situation, but in general what one needs to keep in mind is that the dependence of $|f_n(x)-f_n(y)|$ on $|x-y|$ should allow for some kind of summability to work, ideally to show uniform Cauchyness.
Monotone iterate convergence theorem
A generalized theorem ,now, with different conditions : including not requiring contractivity!
Suppose that $f : \mathbb R \to \mathbb R$ is a monotone decreasing function, and extend $f$ to $\mathbb R \cup \{\pm \infty\}$ by $f( \infty) = \lim_{x \to \infty} f(x)$ and likewise $f(-\infty)$. Suppose that $f_n(\infty) - f_n(-\infty) \to 0$ as $n \to \infty$, and furthermore that for some $N$, $f_N(\infty)$ and $f_N(-\infty)$ are finite. Then, $f$ has a unique fixed point $x^*$ and $f_n \to x^*$ uniformly.
Proof : Note that $f_2$ and hence $f_{2n}$ is increasing for all $n$, hence $f_{2n}(-\infty) \leq f_{2n}(x) \leq f_{2n}(\infty)$. The LHS is an increasing bounded sequence (bounded by $f_{2N}(\infty)$), hence goes to a limit. By our assumption, the RHS goes to the same limit which we call $x^*$, and we've proven by the squeeze theorem that $f_{2n}(x) \to x^*$ for all $x$. A similar kind of reasoning (except with the signs reversed) will tell you that $f_{2n+1}(x) \to x^*$ for all $x$. Now, clearly $f(x) = x^*$ and if $f(y) = y$ then $y = x^*$ is clear from the above.
Let's fix an $\epsilon>0$. Let $N \in \mathbb N$ be such that $n>N$ implies $|f_n(-\infty) 0 f_n(\infty)| < \epsilon$. Then, of course for any $x,y$ we have $|f_n(x) - f_n(y)| < \epsilon$. It follows that $f_n$ is uniformly Cauchy, and hence convergent, finishing the argument. $\blacksquare$
Now, apply this to our case. We know that $f(x) = 4^{-x}$ is a decreasing function. We easily find that $f_2(-\infty),f_2(+\infty)$ are finite. We only need to prove, that $f_n(-\infty) + f_n(+\infty) \to 0$ as $n \to \infty$.
However, what are these sequences? Indeed, note that $f(-\infty) = \infty$ , therefore $f_{n+1}(-\infty) = f_n(\infty)$ for all $n$! Therefore, all we need to do is look at $a_n = f_n(+\infty)$, which is the sequence given by $a_1 = 0, a_n = 4^{-a_{n-1}}$, and we need to prove that $a_{n+1} - a_n \to 0$. It will be sufficient to prove that $a_n$ is convergent for this. Consider the sequence $a_{2n}$ and $a_{2n+1}$, they are individually increasing sequences, both bounded between $0$ and $1$ hence convergent, but both have to converge to $\frac 12$, being the only fixed point of $f_{2n}$ (for this, you require the derivative of $f_{2n}$ but it's still quite easier than the Banach FPT approach).
Hence, we are done!$\blacksquare$
Summary :
A proof of the given statement with generalizabilities(!?!) emphasised.
One generalization via FPT.
Comments on each of the individual generalizabilities.
A theorem using monotonicity conditions only.
