Let $T_{k}: C([0,1]) \to C([0,1])$
Where $(T_{k}x)(s)=\int_{0}^{s}k(s,t)x(t)dt$
Show that $T_{k}$ is well-defined.
My work: We of course simply want to show that $T_{k}x \in C([0,1])$
Let $x \in C([0,1])$ and $s \in [0,1]$ arbitrary, I will initially compute $\vert (T_{k}x)(s)-(T_{k}x)(y)\vert$ for any arbitrary $y < s$ and then come up with estimates.
My computation: $\vert (T_{k}x)(s)-(T_{k}x)(y)\vert=\vert\int_{0}^{s}k(s,t)x(t)dt-\int_{0}^{y}k(y,t)x(t)dt\vert=\vert\int_{y}^{s}k(s,t)x(t)dt+\int_{0}^{y}(k(s,t)-k(y,t))x(t)dt\vert\leq \int_{y}^{s}\vert k(s,t)x(t)\vert dt+ \int_{0}^{y}\vert k(s,t)-k(y,t)\vert \vert x(t)\vert dt\leq\sup\limits_{t}\vert k(s,t)\vert\vert x(t)\vert\vert s -y\vert+ \sup\limits_{t}\vert k(s,t)-k(y,t)\vert\vert x(t)\vert\vert y\vert\leq \sup\limits_{t}\vert k(s,t)\vert\vert x(t)\vert\vert s -y\vert+ \sup\limits_{t}\vert k(s,t)-k(y,t)\vert\vert x(t)\vert$
Now, let $\epsilon >0$.
so for the term $\sup\limits_{t}\vert k(s,t)\vert\vert x(t)\vert\vert s -y\vert$ we can select $\delta_{1}:=\frac{\epsilon}{2\sup\limits_{t}\vert k(s,t)\vert\vert x(t)\vert}$
and for the term $\sup\limits_{t}\vert k(s,t)-k(y,t)\vert\vert x(t)\vert$ we can select $\delta_{2}:=\frac{\epsilon}{2\sup\limits_{t}x(t)}$.
Then for any $y \in [0,1]$ so that $\vert s-y\vert<\delta:=\min\{\delta_{1},\delta_{2}\}$ and thus $\vert (T_{k}x)(s)-(T_{k}x)(y)\vert< \epsilon$
Either I am wrong, or there must be a more elegant solution.
The $\delta_2$ isn't what you want. What's useful is the following: Assuming $k\in C([0,1]^2)$, there is some $\delta_2>0$ such that if $x,y\in[0,1]^2$ and $d(x,y)<\delta_2$ (choose your favorite $p$-metric on $[0,1]^2$), then $|k(x)-k(y)|<\frac{\varepsilon}{2\|x\|}$, where $\|x\|=\sup_{t\in[0,1]}|x(t)|$. Then if $|s_2-s_1|<\delta_2$, we have $d((s_1,t),(s_2,t))<\delta_2$ for all $t\in[0,1]$, and thus $$\int_0^{s_1}|k(s_2,t)-k(s_1,t)||x(t)|\ dt<\varepsilon.$$ Then the rest of the proof goes the same.
As far as elegance is concerned, that's a matter of opinion. But results of this nature don't really have (what I would call) elegant proofs.