If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$

Put $l_n(b) =\sum_{k=1}^n a_kb_k$. $l_n$ is linear, continuous and $\lVert l_n\rVert = \left(\sum_{k=1}^na_k^2\right)^{\frac 12}$. For all $b\in \ell^2$ the sequence $\{l_n(b)\}$ is bounded. By the principle of uniform boundedness we get that the sequence $\left\{\lVert l_n\rVert\right\}$ is bounded.

Following the idea left by david as an answer, I'll post a detailed solution.

Let $T_n:\ell^2 \to \mathbb{C}, T_n(c)=\sum_{j=1}^n a_j c_j$. Then clearly $T_n \in (\ell^2)^*$ for every $n$.

I claim that for every $c\in \ell^2$, the limit $\lim_n T_n(c)=\sum_{j=1}^\infty a_jc_j$ exists.

Indeed, we know it does for $c\in \ell^2$ such that $c(n)\geq 0$ for all $n$. But then, for an arbitrary $c\in \ell^2$, we can decompose $c$ as:

$c=(\mbox{Re } c)^+ - (\mbox{Re } c)^- + i\left( (\mbox{Im }c)^+ - (\mbox{Im } c)^-\right)$

which proves the claim.

As a consequence we have that $\sup_n \lVert T_n(c) \rVert <\infty$ for all $c\in \ell^2$, but then by the uniform boundedness principle, $\sup_n \lVert T_n \rVert <\infty$.

Now, $T_n(c) = \langle c, a^{(n)}\rangle_{\ell^2}$ where $a^{(n)}(j)=\begin{cases} a_j & \mbox{if }j\leq n \\ 0 & \mbox{if } j>n \end{cases}$.

By Riesz' representation theorem on Hilbert spaces, we know that $\lVert T_n \rVert = \lVert a^{(n)}\rVert_2= \left( \sum_{j=1}^n a_j^2 \right)^{\frac{1}{2}}$.

To conclude, since $a_n\geq 0$ for all $n$, we have that $\left( \sum_{n=1}^\infty a_n^2 \right)^{\frac{1}{2}} = \sup_n \lVert a^{(n)} \rVert_2 = \sup_n \lVert T_n \rVert < \infty$, and then $\sum_{n=1}^\infty a_n^2 <\infty$.