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?

You need to prove that for every $\varepsilon > 0$ and $x \in [0,1]$ there is a $\delta > 0$ such that $|x-y| < \delta$ implies that $|Tu_n(x) - Tu_n(y)| < \varepsilon$. The kind of upper bound you exhibit is insufficient to do this (assuming from context that $C$ is a positive constant) since for $\varepsilon = C/2$ your upper bound is always bigger than $\varepsilon$.
Instead, assume without loss of generality that $y > x$ and bound
\begin{align*}
|Tu_n(x) - Tu_n(y)| &= \bigg| \int_0^x (a(x,t) - a(y,t)) u_n(t) dt + \int_x^y a(y,t) u_n(t) dt \bigg| \\
& \leq \int_0^x |a(x,t) - a(y,t)| |u_n(t)| dt + \int_x^y |a(y,t) u_n(t)| dt 
\\& \leq \int_0^x |a(x,t) - a(y,t)| M dt  + M \|a\|_\infty |y-x|
\end{align*}
Now $a$ is continuous on a compact set and hence uniformly continuous so there is a $\delta < (2M \|a\|_\infty)^{-1} \varepsilon$ such that $|x-y| < \delta$ implies that $|a(x,t) - a(y,t)| \leq M^{-1} \frac{\varepsilon}{2}$ for every $t$. So for $|x-y| < \delta$,
\begin{align}
|Tu_n(x) - Tu_n(y)| < \int_0^x \frac{\varepsilon}{2} dt + \frac{\varepsilon}{2} \leq \varepsilon
\end{align}
as desired.
