This is how I would do it:
First assume that $\{a_n\}$ does not converge to zero. This means that there exists $\varepsilon>0$ and a subsequence $\{a_{n_k}\}_k$ with $|a_{n_k}|\geqslant\varepsilon$. Now consider the sequence of vectors $\{e_k\}$, where $e_k$ has a 1 in the $n_k$ position, and zero elsewhere. Then $Te_k$ is the sequence with $a_{n_k}$ in the $n_k$-entry and zeroes elswhere. So $\|Te_k-Te_j\|_2\geqslant\sqrt2\varepsilon$; considering the balls of radius $\varepsilon/2$ centered on the $Te_k$, we produce an infinite number of disjoint balls in $\overline{T(B_X)}$, which shows that $\overline{T(B_X)}$ is not compact, i.e. $T$ is not compact. This proves that if $T$ is compact, then the sequence goes to zero.
Now assume that $\lim a_n=0$. Let $y_1,y_2,\ldots$ be a sequence in $\overline{T(B_X)}$. Fix $\varepsilon>0$. Then we can get a sequence $x_1,x_2,\dots$ in $B_X$ with $\|y_j-Tx_j\|_2<2^{-j}\varepsilon$ for all $j$. Fix $n_0$ such that $|a_n|<\sqrt{\varepsilon/8}$ when $n\geqslant n_0$. Now, for each $k=1,\ldots,n_0$, consider the sequence of $k^{\rm th}$ entries of the sequence $\{x_j\}_j$. As this is a finite number of sequences in the unit ball of $\mathbb{C}$, there is a subsequence $\{x_{j_h}\}_h$ such that its first $n_0$ entries converge. So we can find $h$ such that, for $\ell=1,\ldots,n_0$,
$$
|x_{j_{h+m}}(\ell)-x_{j_h}(\ell)|<\frac{\sqrt\varepsilon}{2^{(\ell+1)/2}K^{1/2}}\ \ \ \text{ for all }m
$$
(i.e. $\{x_{j_h}\}$ is Cauchy in its first $n_0$ coordinates).
Then
$$
\|Tx_{j_{h+m}}-Tx_{j_h}\|_2^2=\sum_{\ell=1}^{n_0}|a_\ell(x_{j_{h+m}}(\ell)-x_{j_h}(\ell))|^2
+\sum_{\ell=n_0+1}^\infty|a_\ell(x_{j_{h+m}}(\ell)-x_{j_h}(\ell))|^2 \\
\leqslant\frac{\varepsilon}2+\frac{\varepsilon}8\,\|x_{j_{h+1}}-x_{j_h}\|_2^2
\leqslant\frac{\varepsilon}2+\frac{\varepsilon}8\,2^2=\varepsilon.
$$
We have shown that $\{Tx_{j_h}\}_h$ is Cauchy, and so it is convergent in $\overline{T(B_X)}$. The sequence $\{y_{j_h}\}_h$ gets arbitrarily close to this sequence, so it is also convergent. So $T$ is compact.
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.
Best Answer
You can use the property of the compact operators: that they map bounded sequences in X to sequences in Y with convergent subsequences. So you can choose your sequence to be an orthonormal basis in your Hilbert space and then you can advance with your proof.