I've shown that if for a sequence $\{f_{n}\}_{n=1}^{\infty}$ in a Hilbert space $H$ we have $$\sum_{n=1}^{\infty}|\langle f,f_n\rangle|^{2}< \infty$$ for all $f\in H$ (i.e., it is a Bessel sequence in $H$), then the map $T$ from $\mathcal{l}^2$ to $H$ sending $(a_n)$ to $\Sigma_n a_n f_n$ is well-defined, linear and bounded.
But I'd appreciate it if someone could guide me on how to prove the converse?
Let $ \varphi: \mathcal{H} \to \mathbb{C} $ be any (continuous) linear functional on $ \mathcal{H} $. Then the composition $$ \varphi \circ T = \left\{ \begin{matrix} {\ell^{2}}(\mathbb{N}) & \to & \mathbb{C} \\ (a_{n})_{n \in \mathbb{N}} & \mapsto & \sum_{n \in \mathbb{N}} a_{n} \varphi(f_{n}) \end{matrix} \right\} $$ is a (continuous) linear functional on $ {\ell^{2}}(\mathbb{N}) $. It follows from the Riesz-Fréchet Theorem that $$ (\spadesuit) \qquad (\varphi(f_{n}))_{n \in \mathbb{N}} \in {\ell^{2}}(\mathbb{N}). $$ Now, pick any $ f \in \mathcal{H} $ and let $ \varphi \stackrel{\text{df}}{=} \langle f,\bullet \rangle_{\mathcal{H}}: \mathcal{H} \to \mathbb{C} $. Then $ \varphi $ is a (continuous) linear functional on $ \mathcal{H} $, and so by $ (\spadesuit) $, we obtain $$ (\langle f,f_{n} \rangle_{\mathcal{H}})_{n \in \mathbb{N}} \in {\ell^{2}}(\mathbb{N}), $$ which is precisely the desired result.