Prove : If $\sum_na_nb_n$ converges whenever $\sum b_n^2 \lt \infty,$ then $\sum a_n^2<\infty$

EDIT: As pointed out in the comments, I was using a slightly different method. Rather, I was proving the map $(b_n)\mapsto(a_nb_n)$ is a bounded linear map from $\ell^2\to\ell^1$. The conclusion of course remains the same, but the method is not quite what the OP wanted.

It's possible to approach the question this way, although it is not the best method. We want to show if $b^n\to0$ in $\ell^2$ and $T(b^n)\to y$ in $\ell^1$ then $y=0$, so it is sufficient to show that $y_k=0$ for each $k$. Fix $\varepsilon>0$. Let $n_0$ be large enough that $\max\{\|b^n\|_2,\|T(b^n)-y\|_1\}<\varepsilon$ for all $n\ge n_0$. Observe that $$|a_kb^n(k)-y_k|\le\|T(b^n)-y\|_1<\varepsilon,$$ so if we can show that $(a_n)$ is a bounded sequence it will follow that $$|y_k|\le\varepsilon+|a_k|\|b^n\|_2<(1+M)\varepsilon$$ where $M=\sup_n|a_n|$, and so we will be done. Suppose for a contradiction that $(a_n)$ is not bounded. Then there is a subsequence $(n_k)$ such that $|a_{n_k}|\ge k$. Let $b=(b_n)$ be the sequence defined by

$$b_n=\begin{cases} \frac1k&\text{if }n=n_k,\\ 0&\text{otherwise.} \end{cases}$$

Clearly $b\in\ell^2$, but $|a_{n_k}b_{n_k}|\ge1$ for all $k$, so $a_nb_n\not\to0$ and in particular $\sum_na_nb_n$ cannot converge. This completes the proof.

You may avoid using the Uniform Boundedness Principle, but lack of continuity of $T$ means that you need to get another control of taking limits. In the present case, you may use monotone convergence, which ultimately relies upon the lattice structure of $\ell^2$. This is also implicit in solutions in previous posts done by hand (without Baire) although perhaps not explicitly stated.

We may assume that every $a_n\geq 0$ (this was in fact done in a previous post, and possibly with the intention of providing a hint, I leave it to you to reduce the problem to this case).

Let $K=\ell^2_+({\Bbb N}) = \{ \beta =(b_n) \in \ell^2: b_n\geq 0\}$ be the positive cone in $\ell^2$

For every $\beta\in K$ our assumption is that $0\leq T\beta= \sum_n a_n b_n < +\infty$. As you mention it suffices to show that $T$ is a bounded linear functional. So suppose it is not. Then you may find a sequence of vectors $\beta_1,\beta_2,...$ in $K$ with the properties that for all $j\geq 1$:

(0) : $\|\beta_j\| \leq 1$

(1) : $T\beta_{j} = 4^j$

Now define the ($\ell^2$-convergent) sum $x = \sum_{j\geq 1} 2^{-j} \beta_j \in K$. Every $\beta_j$ is a positive sequence, so when calculating $Tx$ we only add positive numbers. By monotonicity, we have for every $j\geq 1$:

$$ Tx \geq 2^{-j} T\beta_j = 2^j$$ So $Tx=+\infty$ contrary to hypothesis.

For each $n$, define $a^{(n)}=\{a_{1},a_{2},\ldots,a_{n},0,0,\ldots\}\in l^{2}$. Define $\theta^{n}\in(l^{2})^{\ast}$ by $\theta^{n}(x)=\langle a^{(n)},x\rangle=\sum_{k=1}^{n}a_{k}x_{k}$.

For each $x\in l^{2}$, the sequence $\{\theta^{n}(x)\}_{n}$ is bounded because the infinite series $\sum_{k=1}^{\infty}a_{k}x_{k}$ is convergent. By Uniform Boundedness Principle, $\sup_{n}||\theta^{n}||=M<\infty$. By Riese Representation Theorem, $\sum_{k=1}^{n}a_{k}^{2}=||a^{(n)}||^{2}=||\theta^{n}||^2\leq M^2$. It follows that $\sum_{k=1}^{\infty}a_{k}^{2}<\infty$.