Proof. For each $n \in \natp$, let $A_{n} = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_{n}$ is closed with $\bigcup_{n \in \natp}A_{n} = E$. By the Baire Category Theorem, there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F}< \infty$.

Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)}\subset U$, then for any $y \in E$ with $\norm{y}_{E} \le r$ and $T \in \mathcal{T}$,

so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)}\le 2n/r$.$\square$