Assume $H$ is a separable Hilbert space and $\{f_k\}_{k=1}^\infty$ is a sequence in $H$ such that :
$$\forall f\in H : \sum_{k=1}^\infty \left|\langle f,f_k\rangle\right|^2 < \infty$$
Why $\{f_k\}_{k=1}^\infty$ is then a Bessel sequence ?
That means that $$\exists B>0 : \forall f\in H : \sum_{k=1}^\infty \left|\langle f,f_k\rangle\right|^2 < B\|f\|^2$$
On
As noted, the idea is to show that $T : H \to \ell^2$ given by $Tx = \big(\langle x, f_k\rangle\big)_k$ is bounded.
As an alternative to the uniform boundedness principle, you can also use the closed graph theorem.
Assume $x_n \xrightarrow{n\to\infty} x$ and $Tx_n \xrightarrow{n\to\infty} y \in \ell^2$.
Since $x_n \xrightarrow{n\to\infty} x$, in particular we have $\langle x_n, f_k\rangle \xrightarrow{n\to\infty} \langle x, f_k\rangle$ for all $k \in \mathbb{N}$ because the inner product is continuous.
On the other hand, $Tx_n \xrightarrow{n\to\infty} y$ in $\ell^2$ in particular implies convergence in every coordinate, so for every $k \in \mathbb{N}$ we have $$\langle x_n, f_k\rangle = (Tx_n)_k \xrightarrow{n\to\infty} y_k$$
Therefore, by the uniqueness of the limit we have $y_k = \langle x, f_k\rangle, \forall k \in\mathbb{N}$ so $y = Tx$. Hence, the graph of $T$ is closed so we conclude that $T$ is bounded.
Finally
$$\sum_{n=1}^\infty \left|\langle x, f_k\rangle\right|^2 = \|Tx\|_2^2 \le \|T\|^2\|x\|^2, \forall x \in H$$
so $(f_k)_k$ is a Bessel sequence.
The idea is the following: you consider the operators $T_n:H\to \ell^2$, $n\in \mathbb{N}$ defined as $$(T_nf)_k= \begin{cases}\left \langle f,f_k\right\rangle & k\leq n\\ 0 & k>n \end{cases} $$ Clearly each $T_n$ is a bounded linear operator. From the condition $\sum_{k=1}^{\infty}|\left\langle f,f_k\right\rangle|^2<\infty$ for all $f$ it follows that $T_n\to T$ in the strong operator topology, i.e. $T_nf\to Tf$ for all $f\in H$, where $T:H\to \ell^2$ is the linear operator given by $$Tf = (\left\langle f,f_k\right\rangle)_{k\in \mathbb{N}} $$ It then follows from the uniform boundedness principle that $T$ is bounded as well, which implies the thesis since $\|Tf\|^2=\sum_{k=1}^{\infty}|\left \langle f,f_k\right\rangle|^2$.