Boundedness of subsets in $B(X,Y)$ 
Let $X,Y$ be Banach space and $\mathcal F\subset B(X,Y)$ a subset. If for all $x\in X$ and $y^*\in Y^*$ we have
$$\sup_{T\in \mathcal F}\langle Tx,y^* \rangle<\infty$$
then $\mathcal F$ is bounded in $B(X,Y)$.

This screams for the Uniform Boundedness Theorem. Is the following argument OK?
We can consider $y^*\mapsto\langle Tx,y^* \rangle$ for $\|y^*\|=1$ and conclude from the UBT that $\sup_{T\in \mathcal F}\underbrace{\sup_{\|y^*\|=1}\langle Tx,y^* \rangle}_{\|Tx\|}<\infty$.
By a similar token, consider $x\mapsto \|Tx\|$. Then with the above fact and the UBT, we have $\sup_{T\in \mathcal F} \|T\|<\infty$.
Something about this argument bugs me but I can't put my finger on it.
 A: Your argument is missing something. You are wanting to conclude that
$$
                 |\langle Tx,y^{\star}\rangle| \le M,\;\;\; T\in\mathcal{F},\|y^{\star}\|_{Y^{\star}} \le 1. \tag{$\dagger$}
$$
That requires knowing that
$$
              \sup_{T\in\mathcal{F},\|y^{\star}\|\le 1}|\langle Tx,y^{\star}\rangle| < \infty,\;\;\; \mbox{ for any fixed } x\in X.
$$
That's not something you're given. You're only given that
$$
    \sup_{T\in\mathcal{F}}|\langle Tx,y^{\star}\rangle| < \infty,\;\;\; \mbox{for any fixed } x\in X,\; y^{\star}\in Y^{\star}.
$$
So there's a step in between that you're missing. To bootstrap from the second condition to the first, you can consider the adjoint maps $T^{\star}$. The second condition gives
$$
     \sup_{T\in\mathcal{F}}|(T^{\star}y^{\star})(x)| < \infty,
\;\;\; \mbox{ for any fixed } x\in X, \; y^{\star}\in Y^{\star}.
$$
The uniform boundedness principle gives
$$
          \sup_{T\in\mathcal{F}}\|T^{\star}y^{\star}\|_{Y^{\star}} < \infty,
  \;\;\; \mbox{ for any fixed } y^{\star} \in Y^{\star}.
$$
Now you can apply uniform boundedness again in order to obtain
$$
               \sup_{T\in\mathcal{F}}\|T^{\star}\|_{B(Y^{\star},X^{\star})}  = M < \infty.
$$
Now you get what you want, which was the first inequality $(\dagger)$:
\begin{align}
       |\langle Tx,y^{\star}\rangle| & = |(T^{\star}y^{\star})(x)| \\
        & \le \|T^{\star}y^{\star}\|_{X^{\star}}\|x\|_{X} \\
        & \le \|T^{\star}\|_{B(Y^{\star},X^{\star})}\|y^{\star}\|_{Y^{\star}}\|x\|_{X} \\
        & \le M\|y^{\star}\|_{Y^{\star}}\|x\|_{X}.
\end{align}
And now your argument goes through:
$$
     \sup_{T\in\mathcal{F}}\|Tx\|_{Y}=\sup_{T\in\mathcal{F}}\sup_{\|y^{\star}\|_{Y^{\star}}\le 1}|\langle Tx,y^{\star}\rangle| \le \sup_{T\in\mathcal{F}} M\|x\|_{X} = M\|x\|_{X}.
$$
One more application of uniform boundedness gives $\sup_{T\in\mathcal{F}}\|T\|_{B(X,Y)} \le M$.
