Showing an integral operator is compact Let $K\in L^2(\mathbb{R}^2)$, and define the operator $T: L^2(\mathbb{R})\to L^2(\mathbb{R})$ by 
$$(Tf)(x) = \int K(x, y) f(y) dy$$
I want to show this operator is compact (I imagine this is a standard result). 
I found a proof (I imagine the standard one) of this using that this operator is the limit of finite rank operators. I had another idea though, and I was wondering if it could be fleshed out: 
Let $B$ be a collection of functions in $L^2$ with norm $\leq 1$. The function $G(x) = \lvert \lvert K(x, \cdot) \rvert\rvert_2$ is in $L^2$, and for all $f\in B$, we have $\lvert Tf\rvert \leq G$. So all functions in $T(B)$, which we want to have compact closure, fit below this one $L^2$ function. 
Given a sequence of functions in $T(B)$, if we approximated them in $L^2$ with continuous functions and then found a uniform-on-compact-sets limit of these functions, we could prove the theorem. But for this of course, per Arzela-Ascoli, we need equicontinuity.
Towards proving this, let $L_a: L^2 \to L^2$ be the shift operator $f\mapsto f\circ (x\mapsto x + a)$. We can show that 
$$\lvert \lvert Tf - L_a\circ Tf\rvert \rvert_2 \to 0$$
for all $f\in B$ uniformly. This certainly smacks of equicontinuity. 
My question is: from this above assumption, that all functions in $T(B)$ have this "$L^2$ equicontinuity", can we find continuous functions approximating them to arbitrary precision which are also equicontinuous?
 A: I think I've got it. For each $f\in T(B)$, let $f_a(x) = \frac{1}{a}\int_{x-a/2}^{x + a/2} f(y)dy$. For any $a$, the set $\{f_a: f\in T(B)\}$ is equicontinuous, because by Cauchy-Schwarz,
$$\lvert f_a(y) - f_a(z) \leq \frac{1}{a}\int \lvert 1_{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} \rvert \lvert f(x) \rvert dx$$
$$\leq \frac{1}{a} \lvert \lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} \rvert \rvert_2 \lvert \lvert f\rvert \rvert_2 \leq \frac{1}{a} \lvert \lvert _{(y-a/2, y + a/2)} - 1_{(z-a/2, z + a/2)} \rvert \rvert_2 \lvert \lvert G\rvert \rvert_2$$
which goes to 0 uniformly as $|x- y|\to 0$ for all $f$.
Now see that $f_a(x) \to f$ in $L^2$ as $a\to 0$ for all $f$. Indeed, again in the second line using Cauchy-Schwarz ($(\int_A f d\mu)^2 \leq \mu(A) \int f^2$), and then Tonelli's theorem,
$$\lvert \lvert f_a - f\rvert \rvert_2^2 = \int \lvert \left(\frac{1}{a}\int_{x-a/2}^{x + a/2} f(y)dy\right) - f(x)\rvert^2dx = \int\lvert \frac{\int_{x-1/2}^{x+a/2} (f(y) - f(x))dy}{a} \rvert^2dx$$
$$\leq \int\frac{a\int_{x-1/2}^{x+a/2} \lvert f(y) - f(x)\rvert^2dy}{a^2} dx =\frac{1}{a} \int \int 1_{\{x-a/2 < y < x + a/2\}} |f(x) - f(y)|^2dxdy$$
$$ = \frac{1}{a} \int_0^a \int |f(x + \delta) - f(x)|^2dx d\delta$$
If now $a$ is small enough that $\lvert\lvert f - L_\delta\circ f\rvert \rvert_2 < \epsilon$ whenever $\delta < a$, we have that $\lvert \lvert f_a - f\rvert\rvert_2^2 < \epsilon$. 
These things being established, let's define a sequence in a metric space to be $\epsilon$-Cauchy if eventually its terms differ by less than $\epsilon$. Also, by the arguments so far it is clear that $G_a$ is $L^2$, at least for small $a$, besides being continuous. 
Now, given a sequence $f_n \in T(B)$, let let $a$ be such that $\lvert\lvert(f_n)_a-f_n\rvert\rvert_2 < \epsilon$ for all $n$. The sequence $(f_n)_a$ is equicontinous and uniformly bounded above by a continuous function $G_a$, so on every compact set it by Arzela-Ascoli has a uniformly continuous subsequence. It is not hard to see then that $(f_n)_a$ has an $L^2$-convergent subsequence. As soon this subsequence's terms differ by less than $\epsilon$ (in $L^2$ now), then we have that the corresponding subsequence of $f_n$ is $2\epsilon$-Cauchy. 
Passing to a further subsequence and so on by a diagonalization argument, we get a subsequence of $f_n$ which is $\epsilon$-Cauchy for every $\epsilon$, and then Cauchy, and thus convergent by the completeness of $L^2$. 
