Let $X$ and $Y$ be real Banach spaces and $T:X\to Y$ be linear. Assume that $T$ is $(\omega_X-\omega_Y)-$continuous, then $T$ is $(s_X-s_Y)-$continuous (where $\omega$ denotes 'weak' and $s$ denotes 'strong').
Proof
It suffices to show that Graph of $T$, $G(T)$, is strongly closed. Since the weak topology is Hausdorff and $T$ is $(\omega_X-\omega_Y)-$continuous, then $G(T)$ is closed w.r.t. $\omega_X\times \omega_Y$. This implies that $G(T)$ is $(\omega_{X\times Y})-$closed. Thus, $G(T)$ is $(s_{X\times Y})-$closed.
Since $T\to \Bbb{R}$ is linear and $G(T)$ is $(s_{X\times Y})-$closed, then $T$ is $(s_X-s_Y)-$continuous.
Question
How is $\omega_{X\times Y}=\omega_X\times \omega_Y$. Can someone explain to me, please?
This is not an answer to you question but I would like to suggest use of Uniform Boundedness Principle instead of Closed Graph Theorem. For $y^{*} \in Y^{*}$ with $\|y^{*}\| \leq 1$ define $S_{y^{*}} =y^{*}(Tx)$. It is easy to see that this is a family of continuous linear functionals on $X$ which are bounded at each point $x$. Uniform Boundedness Principle now tells you that $T$ is a bounded operator.