Here is Prob. 7, Sec. 28, in the book Topology by James R. Munkres, 2nd edition:
Let $(X, d)$ be a metric space. If $f$ satisfies the condition $$ d\big( f(x), f(y) \big) < d(x, y) $$ for all $x, y \in X$ with $x \neq y$, then $f$ is called a shrinking map. If there is a number $\alpha < 1$ such that $$ d \big( f(x), f(y) \big) \leq \alpha d(x, y) $$ for all $x, y \in X$, then $f$ is called a contraction. A fixed point of $f$ is a point $x$ such that $f(x) = x$.
(a) If $f$ is a contraction and $X$ is compact, show $f$ has a unique fixed point. [Hint: Define $f^1 = f$ and $f^{n+1} = f \circ f^n$. Consider the intersection $A$ of the sets $A_n = f^n(X)$.]
(b) Show more generally that if $f$ is a shrinking map and $X$ is compact, then $f$ has a unique fixed point. [Hint: Let $A$ be as before. Given $x \in A$, choose $x_n$ so that $x = f^{n+1}\left(x_n\right)$. If $a$ is the limit of some subsequence of the sequence $y_n = f^n \left( x_n \right)$, show that $a \in A$ and $f(a) = x$. Conclude that $A = f(A)$, so that $\mathrm{diam}\, A = 0$.]
(c) Let $X = [0, 1]$. Show that $f(x) = x - x^2/2$ maps $X$ into $X$ and is a shrinking map that is not a contraction. [Hint: Use the mean-value theorem of calculus.]
(d) The result in (a) holds if $X$ is a complete metric space, such as $\mathbb{R}$; see the exercises of \Sec. 43. The result in (b) does not: Show that the map $f \colon \mathbb{R} \to \mathbb{R}$ given by $f(x) = \left[ x + \left( x^2 + 1 \right)^{1/2} \right]/2$ is a shrinking map that is not a contraction and has no fixed point.
Here is my Math SE post on Prob. 7 (a).
And, here is my Math SE post on Prob. 7 (b).
Here I'll only be attempting Part (c).
My Attempt:
Part (c):
Here $$ f(x) = x - \frac{x^2}{2} \ \mbox{ for all } \ x \in [0, 1]. \tag{A} $$ So we obtain $$ f^\prime(x) = 1 - x > 0 \ \mbox{ for all } \ x \in (0, 1). \tag{B} $$ Thus $f$ is a strictly increasing function on $[0, 1]$.
Suppose that there exists a real number $\alpha$ such that $$ 0 \leq \alpha < 1 $$ and such that $$ \big\lvert f(x) - f(y) \big\rvert \leq \alpha \, \lvert x - y \rvert \ \mbox{ for all } \ x, y \in [0, 1]. \tag{0} $$ So if $0 \leq x < y \leq 1$, then we obtain $$ \frac{ \big\lvert f(x) - f(y) \big\rvert }{ \lvert x- y \rvert } \leq \alpha , $$ that is, $$ \alpha \geq \frac{ f(y) - f(x) }{ y - x } = \frac{ \left( y - \frac{y^2}{2} \right) - \left( x - \frac{x^2}{2} \right) }{ y-x } = \frac{ 2(y-x) - \left( y^2 - x^2 \right) }{ 2(y-x) }, $$ Therefore $$ \alpha \geq 1 - \frac{ y+x}{2} \tag{1} $$ for any real numbers $x$ and $y$ such that $0 \leq x < y \leq 1$.
Let $x$ and $y$ be any (fixed but arbitrary) real numbers such that $$ 0 \leq x < y \leq 1. \tag{2} $$ Then by the Lagrange's mean-value theorem there exists a real number $c \in (x, y)$ such that $$ f(y) - f(x) = f^\prime (c) (y-x), $$ and so $$ \frac{ f(y) - f(x) }{ y- x } = f^\prime(c) = 1-c \tag{3}. $$
So using (1) and (3) together, we obtain $$ \alpha \geq 1- c, $$ and hence $$ c \geq 1 - \alpha. \tag{4} $$
However, for any real numbers $x$ and $y$ such that $0 \leq x < y \leq 1$, we obtain $$ \frac{ f(y) - f(x) }{ y-x } = \frac{ \left( y - \frac{y^2 }{2} \right) - \left( x - \frac{x^2}{2} \right) }{ y-x } = \frac{ 2(y-x) - \left( y^2 - x^2 \right) }{2 (y-x) } = 1 - \frac{y + x}{2}, $$ and this together with (3) above yields $$ 1 - \frac{y + x}{2} = 1 - c, $$ and so $$ c = \frac{y + x}{2}. \tag{5} $$
From (4) and (5) above, we obtain $$ \frac{ x+ y}{2} \geq 1- \alpha, $$ and so $$ \alpha \geq 1 - \frac{ x+y}{2} \tag{6} $$ for any real numbers $x$ and $y$ such that $0 \leq x < y \leq 1$.
But if we put $x = 0$ and choose $y$ so that $$ 0 < y < \min \big\{ \, 1, 2 ( 1- \alpha) \, \big\}, $$ then $$0 < y < 2(1-\alpha), $$ which implies $$ 0 < \frac{y}{2} < 1-\alpha, $$ and hence $$ \alpha < 1 - \frac{y}{2} = 1-\frac{x+y}{2}, $$ which contradicts (6) above.
Thus our supposition that there exists a real number $\alpha \in [0, 1)$ for which (0) above holds is wrong. Hence $f$ is not a contraction.
However, if $0 < \leq x < y \leq 1$, then there exists a real number $c \in (0, 1) \subset [0, 1]$ such that $$ f(y) - f(x) = f^\prime(c) (y-x), $$ which implies $$ \big\lvert f(x) - f(y) \big\rvert = f(y)-f(x) = (1-c)(y-x) = (1-c) \lvert x-y \rvert < \lvert x-y \rvert, $$ which shows that $f$ is a shrinking map.
Is this proof good enough? Or, are there issues?
