I'm trying to prove the Uniform Boundness Theorem in Functional Analysis. Suppose $V$ is a Banach Space, and $W$ is a normed space. I want to show that, given a sequence $(T_n)$ of bounded linear operators from $V$ to $W$, if $\forall b \in B:\sup_n ||T_nb|| < \infty$ then $\sup_n||T_n|| < \infty$.
My attempt: Define $C_k = \{b \in B|\sup_n||T_nb|| \leq k \}$. $C_k$ is closed, since if we have a converging sequence $(b_n) \subset C_k,b_n \rightarrow b$, then $\forall n \in \mathbb{N}:||T_nb|| = ||T_n(\lim_{m \rightarrow \infty}b_m)|| = \lim_{m \rightarrow \infty}||T_nb_m|| \leq k \Rightarrow \sup_n ||T_nb|| \leq k \Rightarrow b \in C_k$, since each $T_n$ is continuous.
Also, since we assume that $\sup_n ||T_nb|| < \infty$, we have that $B = \cup_{k \in \mathbb{N}}C_k$. By Baire's Theorem, since $B$ is a banach space, there exists a $C_k$ with an interior point $b_0$, i.e., $\exists \delta: B(b_0;\delta) \subset C_k$, where $B(b_0;\delta)$ denotes the open ball of radius $\delta$ centered at $b_0$.
Then, given any $b \in B:||b|| = 1$, let $\alpha = \delta/2$ and then note that $\alpha b \in B(0;\delta) \Rightarrow \alpha b + b_0 \in B(b_0;\delta)$, which implies that $||T_n(\alpha b + b_0)|| \leq k$.
Thus, $||T_n \alpha b|| = ||T_n(\alpha b + b_0) - T_n b_0|| \leq ||T_n(\alpha b + b_0)|| + ||-T_nb_0|| \leq 2k \Rightarrow \forall n \in N:||T_n b || \leq 2k/\alpha$. Then, $\forall n \in N:||T_n|| = \sup_{||b||=1}||T_nb|| \leq 2k/\alpha$, hence $\sup_n||T_n|| \leq 2k/\alpha \Rightarrow \sup_n||T_n|| < \infty$.
Is this proof correct? The proof that my textbook provides is essentially the same, the only difference beeing that $C_k = \{b \in B|\sup_n||T_nb|| \leq k, ||b|| \leq 1 \}$. Do we really need to ensure that $||b|| \leq 1$?