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.
Best Answer
This may come as a surprise but the answer is no. In fact it is no in a very strong way!!
There exists a sequence $\{T_n\}_n$ of bounded operators, strongly converging to zero, such that $\{KT_n\}_n$ does not converge to zero in norm for every nonzero bounded operator $K$, regardless of whether $K$ is compact or even finite rank!
Let $S:\ell ^2\to \ell ^2$ be the so called unilateral shift given by $$ S(x_0, x_1, x_2, x_3, \ldots ) = (0, x_0, x_1, x_2, \ldots ). $$ The adjoint of $S$ turns out to be given by $$ S^*(x_0, x_1, x_2, \ldots ) = (x_1, x_2, x_3, \ldots ), $$ from where one has that $S^*S$ is the identity operator.
Now let $T_n={S^*}^n$, so that $$ T_n(x_0, x_1, x_2, \ldots ) = (x_n, x_{n+1}, x_{n+2}, \ldots ), $$ and hence it is clear that $T_n\to 0$ strongly.
If $K$ is any bounded operator (compact or not) we have that $$ \|K\| = \|K{S^n}^*S^n\| \leq \|K{S^n}^*\|\|S^n\| = \|K{S^n}^*\| = \|KT_n\|, $$ so we can't have $\|KT_n\|\to 0$, unless $K=0$.
As mentioned in the original post, if $K$ is compact, it is well known that "$T_n\to 0$ strongly" implies "$T_nK\to 0$ in norm".
If $T_n^*\to 0$ strongly (e.g. if $T_n\to 0$ in the $^*$-strong topology, or if $T_n$ is self-adjoint and converges to zero strongly) then $T_n^*K^*\to 0$ in norm by (1) and hence $KT_n\to 0$ in norm as well.