Compact operator by proving Ascoli-Arzelà I need to prove that this operator satisfies Ascoli-Arzelà's hypothesis.

$T: C^0[0,1] \rightarrow C^0[0,1] $, defined $Tu(x)=\int_0^x a(x,t) u(t)dt$, where $a(x,t)=C^0 ([0,1] \times [0,1])$.

Equiboundedness is okay, I need to prove equicontinuity: taken $\{ u_n \} \subset C^0[0,1], ||u_n|| \leq M$
So I take $|Tu_n(x)-Tu_n(y)| \leq \ldots \leq L|x-y| + C$ , by bounding several times. Is it enouth to conclude?

If you have a nice strategy to induce (uniformly) bounded $L$ and sufficiently small $C>0$ (depending on $\epsilon$, of course), then it will work. But I suspect that it cannot be implemented in most cases. You don't have to get a Lipschitz-like, very concrete bound to show equicontinuity. Here's an approach.
Let $M$ be a constant such that $\sup_n \|u_n\|\le M$ and $|\alpha|\le M$, and define$$
m(\delta)=\sup\{|\alpha(t,x)-\alpha(t',x')|:|(t-t',x-x')|\le \delta\}.
$$ By continuity of $\alpha$, we have $\lim\limits_{\delta\to 0^+}m(\delta)=0.$ Let $\epsilon>0$ be given and we can find $\delta>0$ such that $m(\delta)\le \epsilon/M$. Note that we can choose $\delta$ so that $\delta<\epsilon/M^2$. We find that for all $0\le x- y\le \delta$ and $n\ge 1$,
$$\begin{eqnarray}
|Tu_n(x)-Tu_n(y)|&\le&\int_0^y |\alpha(x,t)-\alpha(y,t)||u_n(t)|\mathrm{d}t+\int_y^x |\alpha(x,t)||u_n(t)|\mathrm{d}t\\&\le& Mm(\delta)+M^2 \delta <2\epsilon.
\end{eqnarray}$$ This verifies equicontinuity of $(Tu_n)$ as desired.
