Bessel Sequence proof check. I have a similar question to Why can Bessel sequences be defined by the condition $\sum_{k=1}^{\infty}|\langle f,f_{k}\rangle|^{2}<\infty$?, where it was solved using Banach-Steinhaus.
A sequence $\{f_k\}_{k=1}^{\infty}$ is called a Bessel sequence in a Hilbert space $\mathscr{H}$, if $\sum_{k=1}^{\infty}|\langle f,f_k\rangle|^2 < \infty \, \forall f \in \mathscr{H}$.
Statement: Given a Bessel sequence use the close graphing theorem to show that  $\sum_{k=1}^{\infty}|\langle f,f_k \rangle|^2 \leq B\|f\|^2$ where $B < \infty$
Attempted proof: Let $\{f_i\}_{i=1}^{\infty} \, \in \, \mathscr{H}$ where $\lim_{i\rightarrow \infty} f_i = f$ and define $T(f) = \{\langle f, f_n\rangle\} = \{F_n\}_{n=1}^{\infty}\, \in \, l_2$
Now for a fixed $m$ we have:
$|F_m - \langle f_i, f_m\rangle| \leq |\sum_{k=1}^{\infty}|F_k -\langle f_i,f_k\rangle|^2|^{\frac{1}{2}} = \|{F_k} - T(f_i)\|_2 \rightarrow 0 \, \mbox{as} \, i\rightarrow \infty$
hence
$\lim_{i\rightarrow \infty} \langle f_i, f_m\rangle = F_m$
So we have convergence hence we are closed and by the CGT
$\sum_{k=1}^{\infty}|\langle f,f_k \rangle|^2 \leq B\|h\|^2$ where $B < \infty$
Q.E.D.
 A: It looks like you're using the Banach-Steinhaus Theorem, not the Closed Graph Theorem.
First note that since $(f_i)$ is a Bessel sequence, the map $T$ is well-defined.
Clearly, $T$ is linear. We wish to show that $T$ is bounded,  which  is precisely what the  inequality in your Statement says.
To use the Closed Graph Theorem, you can show the following holds:

$\ \ \ $Suppose $(x_n)$ is a sequence in $H$, $x_n\rightarrow x$, and $Tx_n\rightarrow y$, where $x\in H$ and $y\in\ell_2$.  
$\ \ \ $Then $y=Tx$.

If you can do this, the Closed Graph Theorem will tell you $T$ is bounded;
So suppose  $(x_n)$ is a sequence in $H$ that converges to $x\in H$, and that $Tx_n\rightarrow y\in \ell_2$. In the following, we denote the $j$'th coordinate of a vector $x$ by $x(j)$.
Fix a coordinate $j$. We'll show $(Tx)(j)=y(j)$. As, $j$ is arbitrary, once we've done this, we can conclude $Tx=y$. 
The $j$'th coordinate of $Tx_n$ is $f_j(x_n)$. Since $x_n\rightarrow x$, as $n\rightarrow\infty$, and since $f_j$ is a continuous linear functional on $H$, we have $$(Tx_n)(j)=f_j(x_n)\ \buildrel{n\rightarrow\infty}\over\longrightarrow \ f_j(x) =(Tx)(j).$$ In short, $Tx_n$ converges to $Tx$ coordinatewise. As mentioned above, this implies $y=Tx$, as desired. 
