The problem:
$X,Y,Z$ are Banach spaces and $B:X\times Y\to Z$ is a bilinear map which is separately continuous. i.e. $\forall x\in X$ the map $B(x,\cdot):Y\to Z$ is continuous, and $\forall y\in Y$ the map $B(\cdot,y)$ is continuous. Then, $B:X\times Y\to Z$ is continuous.
Proof attempt:
It suffices to prove that $B$ is a bounded linear map since if $B$ is bounded, it is also continuous by Proposition 5.2 from the text. Clearly it is linear since it is bilinear. The goal is to show boundedness:
Define $B_r:X\to\mathcal{L}(Y,Z)$ by $B_r(x)=B(x,\cdot)$ and $B_l:Y\to\mathcal{L}(X,Z)$ by $B_l(y)=B(\cdot,y)$. Clearly $B_r(x)(y)=B(x,y)=B_l(y)(x)$. In particular then, $||B(x,y)||=||B_r(x)(y)||\leq||B_r(x)||||y||\leq||B_r||||x||||y||$. So it suffices to show $||B_r||<\infty$. Now $||B_r||=\sup\limits_{x\in X}(||B_r(x)||: ||x||=1)$ by definition. Define $B_1$ to be the ball in $X$ centered at 0 with radius 1. Now by the Uniform Boundedness Principle, if we can show that $\sup\limits_{x\in B_1}(||B_r(x)(y)||)<\infty$ for all $y\in Y$ (since $Y$ is non-meager in itself), it follows $\sup\limits_{x\in B_1}(||B_r(x)||<\infty)$. But $||B_r(x)(y)||=||B_l(y)(x)||\leq ||B_{l}(y)||$ for every $y$. And since for each $y$ it is the case that $B_l(y)$ is an element of $\mathcal{L}(X,Z)$, it is bounded. It follows that $\sup\limits_{x\in B_1}(||B_r(x)||)<\infty$. But this is just $||B_r||$ so the result follows.
I am confused about the last couple of steps in my own proof. In particular, I wonder if I have really satisfied the requirement to invoke the result of the UBP. Help is appreciated. Thanks!