Here's a proof I learned from some notes of Richard Melrose. I just noticed another answer was posted while I was typing. This uses a different characterization of compactness so hopefully it is interesting for that reason.
First, I claim that a set $K\subset H$ of a Hilbert space is compact if and only if it is closed, bounded, and satisfies the equi-small tail condition with respect to any orthonormal basis. This means that given a basis $\{e_k\}_{k=1}^\infty$, for any $\varepsilon>0$ we can choose $N$ large enough such that for any element $u\in H$,
$$\sum_{k>N} |\langle u , e_k \rangle |^2 < \varepsilon.$$
The main point here is that this condition ensures sequential compactness. The proof is a standard "diagonalization" argument where you choose a bunch of subsequences and take the diagonal. This is done in detail on on page 77 of the notes I linked.
With this lemma in hand, the proof is straightforward. I repeat it from page 80 of those notes. Fix a compact operator $T$. By definition [this is where we use a certain characterization of compactness] the image of the unit ball $T(B(0,1))$ is compact. Then we have the tail condition that, for given $\varepsilon$, there exists $N$ such that
$$\sum_{k>N} |\langle Tu , e_k \rangle |^2 < \varepsilon.$$
for any $u$ such that $\| u \| < 1$.
We consider the finite rank operators
$$T_nu = \sum_{k\le n} \langle Tu , e_k \rangle e_k.$$
Now note the tail condition is exactly what we need to show $T_n \rightarrow T$ in norm. So we're done.
It is well-known, that $G\in L^2((0,1)\times (0,1))$ implies that $A$ is a compact operator. As we are on a separable Hilbert space by, e.g., Lemma 3.3.c in this paper, all that is left to show that $T_n$ converges strongly to the identity (because then $T_nAT_n\to \operatorname{id}A\operatorname{id}=A$ in norm as you wanted).
Now $T_n$ is said to converge strongly to $\operatorname{id}$ if for all $x\in L^2$, one has $\lim_{n\to\infty}\|T_nx-x\|_{L^2}=0$. This however holds in arbitrary separable infinite-dimensional Hilbert spaces as is easily seen using the Fourier expansion and Parseval's identity.
Best Answer
You can't show it because it is false. For instance, consider $X=H$ where $H$ is infinite-dimensional. Then $H \otimes X$ are the finite-rank operators $H \to H$, and their norm-closure is the compact operators.