Is the kernel mapping compact

Answering to the comments: yes, it is most likely that $C(X)$ is meant with the supremum norm.

Let $(u_n)\subset C[0,1]$ be a bounded seqeunce, say there exists $M>0$ s.t. $\|u_n\|_\infty\le M$ for all $n$. We shall show that $Tu_n$ has a convergent subsequence.

Well, what can we use to do that, if not the infamous Arzela-Ascoli theorem?

Note first that, if $x\in [0,1]$, then $|Tu_n(x)|\le\int_0^1 |k(x,s)u_n(s)|ds\le\sup_{r,s\in X}|k(r,s)|\cdot\sup_{s\in X}|u_n(s)|=\|k\|_{C([0,1]^2)}\cdot\|u_n\|\le\|k\|\cdot M$. This shows that $\|Tu_n\|\le\|k\|\cdot M$ and this is true for all $n$, so the family $\{Tu_n\}$ is uniformly bounded.

Now given $r,s\in [0,1]$ we have $$|Tu_n(r)-Tu_n(s)|\le\int_0^1|k(r,t)-k(s,t)|\cdot|u_n(t)|dt\le M\int_0^1|k(r,t)-k(s,t)|dt$$ Since $k$ is continuous on a compact set, it is uniformly continuous; therefore, given $\varepsilon>0$ we can find $\delta:=\delta(\varepsilon)>0$ (depending only on $\varepsilon$ and $k$!) such that $$\|(a,b)-(a',b')\|_{[0,1]^2}<\delta\implies|k(a,b)-k(a',b')|<\varepsilon$$

So assume that $\varepsilon>0$ is given. For the quantity $\varepsilon/M>0$ take the corresponding $\delta$, i.e. take $\delta=\delta(\frac{\varepsilon}{M})$. Now if $|r-s|<\delta$ we have that $$\|(r,t)-(s,t)\|_{[0,1]^2}=\sqrt{|r-s|^2+|t-t|^2}=|r-s|<\delta$$ for any $t\in[0,1]$, so $|k(r,t)-k(s,t)|<\varepsilon/M$ for all $t\in[0,1]$. But our earlier computation shows that $|Tu_n(r)-Tu_n(s)|<\varepsilon$. We have just shown that $\{Tu_n\}$ is equicontinuous.

Well, now that it's applicable, apply Arzela-Ascoli. Wait! Didn't we just verify the definition of a compact operator?