So there's one proof that uses $X,Y$ normed spaces $T \in Lin_b(X,Y)$.
The draws a $g \in Y^*$.
This part of the proof concerns showing that $g(Tx_n)\rightarrow g(Tx)$.
Why is it enough to consider only $||g||_{Y^*}=1$.
Is it because if the norm was anything different, then one could always normalize it?
Yes, precisely. Suppose $||g||_{Y^*} = k$, then we can take $h = \frac{g}{k}$ which is also in $Y^*$. Then, the given argument would show $h(T x_n) \to h(T x)$, but then $k h(T x_n) \to k h(T x)$ since we have normed spaces. Hence, $g(T x_n) \to g(T x)$, as required.