Every bounded sequence of dual space contains a subsequence which is weak* convergent

The idea is to diagonalize, as mentioned earlier, but you have to do it carefully: Let $\{x_n\}$ be a countable dense subset of $X$. Not $\{T_n(x_1)\}$ has a convergent subsequence by Bolzano-Weierstrass, which you index with an increasing sequence $\{s(1,n) : n\in \mathbb{N}\} \subset \mathbb{N}$.

Now $\{T_{s(1,n)}(x_2)\}$ has a convergent subsequence, which you index by an increasing $\{s(2,n) : n\in \mathbb{N}\}$. Proceed inductively to obtain strictly increasing sequences $\{s(j,n) : n\in \mathbb{N}\}$ for each $j \in \mathbb{N}$ such thath

• $\{s(j+1,n) : n \in \mathbb{N}\}$ is a subsequence of $\{s(j,n) : n \in \mathbb{N}\}$
• $\{T_{s(j,n)}(x_j)\}$ is convergent for each $j \in \mathbb{N}$

Now consider the subsequence $T_{s(n,n)}$, and this converges pointwise at each $x_j$. As you mention, this completes your proof.