Let $(f_{n})_{n}\subseteq C([0,1])$ be an arbitrary sequence such that $\vert \vert f_{n} \vert \vert_{\infty}\leq 1$ for all $n \in \mathbb N$ and $k \in C([0,1]^{2})$.
Define the bounded linear map $T: (C([0,1]),\vert \vert \cdot \vert \vert_{\infty}) \to (C([0,1]),\vert \vert \cdot \vert \vert_{\infty}), f\mapsto Tf$
where $Tf: [0,1]\to \mathbb K, x\mapsto \int_{0}^{x}k(x,y)f(y)dy$
Show that $(Tf_{n})_{n}$ has a convergent subsequence.
My idea: I want to define a new function $g_{n}:=[0,1]\to \mathbb K, x \to \int_{0}^{x}k(x,y)f_{n}(y)dy$ and show that it is is uniformly continuous. Then I can use Arzela-Ascoli to prove the existence of a subsequence of $(Tf_{n})_{n}$
But I do not know how to go about this. Am I even on the right track?
Call the family of functions $\mathcal{F}$. You can just show that $T\left(\mathcal{F}\right)$ is relatively compact, using the Arzela-Ascoli theorem. To do this, you must show that $T\left(\mathcal{F}\right)$ is pointwise bounded and equicontinuous. For now, I'll leave you hints, and I can fill in details, if you'd like.
Pointwise bounded: you know that $\mathcal{F}$ is bounded. You can deduce that $k$ is bounded, too (why?). Use these facts in a standard bounding argument.
Equicontinuous: The function $k$ is given to be continuous. It's actually uniformly continuous (why?). Use this to get your $\delta$ for equicontinuity.