Prove that this integral operator is compact

To show that $K$ is compact using elementary arguments, let $\{ f_{n} \}_{n=1}^{\infty}$ be a bounded sequence in $L^{2}[0,1]$ with $\|f_{n}\| \le M$. For every $\epsilon > 0$, there exists $\delta > 0$ such that $|k(x,y)-k(x',y')| < \epsilon$ whenever $|x-x'|+|y-y'| < \delta$. Therefore, $\{ Kf_{n}\}$ is a sequence of continuous functions for which $$ |Kf_{n}(x)-Kf_{n}(x')| \le \int_{0}^{1}|k(x,y)-k(x',y)||f_{n}(y)|\,dy \\ \le \epsilon \int_{0}^{1}|f_{n}(y)|\,dy \le \epsilon\|1\|\|f_{n}\| \le M\epsilon,\;\;\; |x-x'| < \delta. $$ That shows that $\{ Kf_{n} \}$ is an equicontinuous family of continuous functions on $[0,1]$. So, there exists a subsequence $\{ Kf_{n_{k}}\}$ that converges uniformly to a continuous function $g$. Since uniform convergence implies convergence in $L^{2}[0,1]$, it follows that $\{ Kf_{n_{k}}\}$ converges in $L^{2}[0,1]$. Therefore $K$ is compact because the image of a bounded sequence always contains a convergent subsequence.

Using Stone-Weierstass theorem, the function $k$ can be approximated uniformly in $[0,1]^2$ by linear combinations of functions of the form $(x,y)\mapsto u(x)v(y)$, where $u,v\colon [0,1]\to \mathbb R$ are continuous functions.

For such functions, the kernel operator has a finite rank. It remains to show that uniform convergence of the kernel implies convergence for the operator norm.

Actually, if $k\in L^2(0,1)^2$, then you may consider $K$ as a limit of a sequence of finite rank operators. Since $L^2(0,1)$ is a separable Hilbert space, there exists a Hilbert basis denoted by $\{e_i\}_{i=1}^{\infty}$. Define $T_n (n\in\mathbb{N}^{+})$ as $$ T_nf:=\sum\limits_{i=1}^{n}\langle f, e_i \rangle e_i,\quad \forall f\in L^{2}(0,1). $$ Then, $K_n:=T_nKT_n$ is a finite rank operator. And $K_n$ is compact for each $n$. You need to show that $\lim\limits_{n\rightarrow \infty}||K_n-K||=0$. Then, you conclude that $K$ is compact, since the space of compact operators mapping $L^2(0,1)$ into $L^2(0,1)$ is closed.