TITLE: Compactness with Ascoli-Arzelà?
QUESTION [5 upvotes]: Let 
  $$
K:= \{x: [0,T] \to \mathbb R: x^{\prime}(t)=x^2(t), \, 0\le x(T) \le 1\}.
$$
  Prove that $K$ is a compact subset of $C([0,T],\mathbb R)$.

My idea is to use Ascoli-Arzelà thm. First of all, by some routine calculations I have found that 
$$
K= \left\{x:[0,T] \to \mathbb R: x(t)=\frac{x(T)}{x(T)(T-t)+1}, \, 0\le x(T) \le 1\right\}
$$
Then I have shown that 

$K$ is equibounded:
$$
\vert x(t)\vert\le\left\vert \frac{x(T)}{(T-t)x(T)+1}\right\vert \le x(T) \le 1
$$
for every $t \in[0,T]$ and for every $x(\cdot) \in K$.
$K$ is equicontinuous: indeed the following holds true
$$
\vert x(t_1)-x(t_2) \vert \le \vert t_1 -t_2 \vert 
$$
for every $t_1,t_2 \in[0,T]$ and for every $x(\cdot) \in K$.

Finally I have to show that $K$ is closed: let us take $x_n \to x$ uniformly s.t. $x^{\prime}_n = x_n^2$ for every $n$. Then $x_n^2 \to x^2$ uniformly; to sum up we have 

$x_n \to x$ uniformly;
$x^{\prime}_n = x_n^2 \to x^2$ uniformly. 

Hence $x$ is differentiable and $x^{\prime}=x^2$. 
Is my proof correct? Thanks in advance.

REPLY [3 votes]: Yes, it's correct. I would only add after the 

let us take $x_n \to x$ uniformly s.t. $x^{\prime}_n = x_n^2$ for every $n$. Then $x_n^2 \to x^2$ uniformly;

the justification that $\{x_n\}$ is uniformly bounded by $1$.