Corollary:
For the corollary notice that for all $x \in [0,1]$
$$
p'(x)=mnx^{n-1}(1-x^n)^{m-1} \geq 0.
$$
So $p$ is non-decreasing in $[0,1]$. Notice that $p(0)=0<p(\frac{c}{2})<\epsilon$. Hence $0 \leq p(x)<\epsilon$ for all $x \in [0,\frac{c}{2}]$. And notice that $p(c) \geq 1- \epsilon$. Hence $p(x) \geq 1- \epsilon$ for all $x \in [c,1]$.
Without using differantiation:
Suppose $x<y \in [0,1]$. Then $x^n<y^n$ and $-x^n>-y^n$. Hence $1-x^n>1-y^n$ and $(1-x^n)^m>(1-y^n)^m$. So $-(1-x^n)^m<-(1-y^n)^m$ and $1-(1-x^n)^m<1-(1-y^n)^m$. This implies that $p(x)<p(y)$ and $p$ is increasing.
Lemma:
This is my attempt on proving the Lemma where I used Bernoulli's inequality which is a standard result and with an easy proof. Hope it helps and if there is something wrong with the proof please let me know.
Edit: Bernoulli's inequality states that
$$
(1+x)^r \geq 1+rx
$$
for any integer $r \geq 0$ and for every real number $x > -1$. For a proof of Bernoulli's inequality take a look at Proof by induction of Bernoulli's inequality $ (1+x)^n \ge 1+nx$. It is a simple application of induction.
Notice that we can suppose $\epsilon<1$. Now, since $0<c<1$, we know that $\frac{1}{c} > 1$ and we can choose $k$ the smallest positive integer such that $k>\frac{1}{c}$. Then we have
$$
k-1 \leq \frac{1}{c} \implies k \leq \frac{1}{c}+1<\frac{2}{c}.
$$
Also notice that $-\frac{c^n}{2^n}> -1$ for any $n \in \mathbb{N}$. Taking $x=-\frac{c^n}{2^n}$ and $r=k^n$ we can apply Bernoulli's inequality
$$
\left(1+\left(-\frac{c^n}{2^n}\right) \right)^{k^n} \geq 1-k^n\frac{c^n}{2^n}=1-\left(k\frac{c}{2} \right)^n
$$
Then follows from $k<\frac{2}{c}$ that $\delta_0=(k\frac{c}{2}) < 1$ and
\begin{equation}
\left(1-\frac{c^n}{2^n} \right)^{k^n} \geq 1-{\delta_0}^n.
\end{equation}
Now let's use Bernoulli's inequality again using that $c^n > -1$ for any $n \in \mathbb{N}$. This time take $x=c^n$ and $r=k^n$ and apply the inequality
$$
(1-c^n)^{k^n} = \frac{1}{(kc)^n}(1-c^n)^{k^n}k^nc^n \leq \frac{1}{(kc)^n}(1-c^n)^{k^n}(1+k^nc^n)
$$
$$
(1-c^n)^{k^n} \leq \frac{1}{(kc)^n}(1-c^n)^{k^n}(1+c^n)^{k^n}=\frac{1}{(kc)^n}(1-c^{2n})^{k^n}<\frac{1}{(kc)^n}.
$$
First equality holds because we multiplied and divided by $k^nc^n$ and first equality holds because $k^nc^n<1+k^nc^n$. Then we just used Bernoulli's inequality on $1+k^nc^n$ and used the fact that $0<1-c^{2n}<1$.
Now using that $k>\frac{1}{c}$ we have $\delta_1=\frac{1}{kc}<1$. Then
\begin{equation}
(1-c^n)^{k^n}<{\delta_1}^n.
\end{equation}
Notice that ${\delta_0}^n \to 0$ and ${\delta_1}^n \to 0$ as $n \to \infty$ (because $0<\delta_0,\delta_1<1$). Hence, using the definition of limit, we can choose $n_0$ such that ${\delta_0}^{n_0}<\epsilon$ and ${\delta_1}^{n_0}<\epsilon$. From what we already proved we have
$$
\left(1-\frac{c^{n_0}}{2^{n_0}} \right)^{k^{n_0}} \geq 1-{\delta_0}^n>1-\epsilon
$$
$$
(1-c^{n_0})^{k^{n_0}} < {\delta_1}^n<\epsilon.
$$
By choosing $n=n_0$ and $m=k^{n_0}$ we have
$$
p \left(\frac{c}{2} \right)=1-\left(1-\frac{c^{n_0}}{2^{n_0}}\right)^{k^{n_0}}<1-(1-\epsilon)=\epsilon
$$
$$
p(c)=1-(1-c^{n_0})^{k^{n_0}}>1-\epsilon.
$$
