Banach-Steinhaus theorem problem [closed]

I don't think that you need the Banach - Steinhaus for this, your condition is strong enough to prove it without it. Let me remind you a nice corollary of the Hahn - Banach:

Let $X$ be a Banach space and $x\in X$. Then there exists $x^*\in X^*$ with $||x^*||=1$ such that $x^*(x) = ||x||$.

Now, fix some $x\in B_X$. Using the above for the image $T(x)$ we may find $g_x \in Y^*$ with $||g_x||=1$ and $g_x(T(x))=||T(x)||$. Then by your hypothesis we know that the set $\{g_x(T(x)):\, x\in B_X\}$ is bounded. But this set is equal to $\{||T(x)||\,:\, x\in B_X\}$ and therefore, it follows that $$||T|| = \sup\{||T(x)||\,:\, x\in B_x\}<\infty.$$

Edit Proof of the equality $$||T|| = \sup\{g(T(x))\,:\, g\in B_{Y^*},\, x\in B_X\}\tag{*}=:M.$$

Let $\epsilon>0$, then by sup's characterization there exists $g\in B_{Y^*},\, x\in B_X$ such that $$M-\epsilon < g(T(x)).$$ Then, $$M-\epsilon<|g(T(x))|\leq ||g||\cdot ||x||\cdot ||T|| = ||T||.$$ Therefore, $M-\epsilon<||T||$ for every $\epsilon>0$, which in turn implies that $M = ||T||$ and $(*)$ has been proven.