[proofplan] We exploit the completeness of $X$ through the Baire Category Theorem. The pointwise hypothesis implies $X = \bigcup_n F_n$ where $F_n = \{x : \sup_\alpha \|T_\alpha x\| \leq n\}$. Baire gives some $F_N$ with nonempty interior containing a ball $B(x_0, r)$. A centering argument then converts this local bound into a uniform operator norm estimate $\sup_\alpha \|T_\alpha\| \leq 2N/r$. [/proofplan] [step:Define the exhaustion and show each level set is closed] For each $n \in \mathbb{N}$, define \begin{align*} F_n = \left\{x \in X : \sup_{\alpha \in A} \|T_\alpha x\|_Y \leq n\right\} = \bigcap_{\alpha \in A} \left\{x : \|T_\alpha x\|_Y \leq n\right\}. \end{align*} Pointwise boundedness gives $X = \bigcup_n F_n$. [claim:Each Level Set Is Closed] $F_n$ is closed for every $n$. [/claim] [proof] For each $\alpha$, $x \mapsto \|T_\alpha x\|_Y$ is continuous, so $\{x : \|T_\alpha x\|_Y \leq n\}$ is closed. $F_n$ is an intersection of closed sets, hence closed. [/proof] [/step] [step:Apply the Baire Category Theorem to find a ball inside some $F_N$] Since $X$ is a Banach space, it is a complete metric space. The Baire Category Theorem applied to $X = \bigcup_n F_n$ (each $F_n$ closed) guarantees that some $F_N$ has nonempty interior. Choose $x_0 \in X$ and $r > 0$ with $B(x_0, r) \subseteq F_N$. [/step] [step:Extract the uniform bound $\sup_\alpha \|T_\alpha\| \leq 2N/r$] [claim:Uniform Operator Norm Bound] $\sup_{\alpha \in A} \|T_\alpha\|_{\mathcal{L}(X, Y)} \leq 2N/r$. [/claim] [proof] Let $\|x\|_X \leq 1$. Then $x_0 + rx \in B(x_0, r) \subseteq F_N$, so $\sup_\alpha \|T_\alpha(x_0 + rx)\|_Y \leq N$. Also $x_0 \in F_N$ gives $\sup_\alpha \|T_\alpha x_0\|_Y \leq N$. By linearity and the triangle inequality: \begin{align*} r\|T_\alpha x\|_Y = \|T_\alpha(x_0 + rx) - T_\alpha x_0\|_Y \leq \|T_\alpha(x_0 + rx)\|_Y + \|T_\alpha x_0\|_Y \leq 2N. \end{align*} Dividing by $r$: $\|T_\alpha x\|_Y \leq 2N/r$ for all $\alpha$ and all $\|x\| \leq 1$. Therefore \begin{align*} \sup_{\alpha \in A} \|T_\alpha\|_{\mathcal{L}(X, Y)} \leq \frac{2N}{r} < \infty. \end{align*} [/proof] [/step]