On separable Hilbert space $H$, weak operator topology is metrizable on bounded parts of $B(H)$ The following is a theorem of Takesaki's operator theory: 

In this proof, weak topology means weak operator topology.
I'm wonder why the theorem holds just for bounded parts of $B(H)$ and also where does he  use of boundedness?  Thanks.
 A: The use of boundedness of $S$ is contained in the "It is not hard to see..." statement. 
Suppose $\{x_i\}_{i\in I}$ is a net in $S$. We have to prove that convergence of the net in the weak operator topology (wot) is equivalent to convergence with respect to $d$.
Assume that $x_i\to x$ in the wot and let $\epsilon>0$. Let $N_0\in\mathbb N$ be such that $\displaystyle2\sum_{n,m>N_0}2^{-(n+m)}<\epsilon$, and $i_0\in I$ such that $\lvert((x-x_i)\xi_n|\xi_m)\rvert<\epsilon$ for all $m,n\leq N_0$ and $i\geq i_0$. Then
\begin{align*}
d(x_i,x)&=\sum_{n,m\leq N_0}2^{-(n+m)}\lvert((x-x_i)\xi_n|\xi_m)\rvert+\sum_{n,m>N_0}2^{-(n+m)}\lvert((x-x_i)\xi_n|\xi_m)\rvert\\
&\leq C\epsilon+\epsilon,
\end{align*}
where $C=\sum_{n,m\leq N_0}2^{-(n+m)}$, and where we have used $\lvert((x_i-x)\xi_n|\xi_m)\rvert\leq2$ which follows since $x_i,x\in S$ and $\lVert\xi_k\rVert\leq1$ for all $k$. Conclusion: $d(x_i,x)\to0$.
Conversely, suppose $d(x_i,x)\to0$. Then clearly $((x-x_i)\xi_n|\xi_m)\to0$ for all $n,m\in\mathbb N$. Now let $\xi,\eta\in\mathfrak H$ and let $\epsilon>0$. By density of the sequence $\{\xi_k\}$ in the unit ball of $\mathfrak H$ we can find $n,m$ such that $\lVert\xi-\xi_m\rVert,\lVert\eta-\xi_n\rVert<\epsilon$. Then
\begin{align*}
\lvert((x-x_i)\xi|\eta)\rvert&=\lvert((x-x_i)\xi|\eta)-((x-x_i)\xi|\xi_n)+((x-x_i)\xi|\xi_n)-{}\\
&\qquad{}-((x-x_i)\xi_m|\xi_n)+((x-x_i)\xi_m|\xi_n)\rvert\\
&\leq\lVert x-x_i\rVert\lVert\xi\rVert\lVert\eta-\xi_n\rVert+\lVert x-x_i\rVert\lVert\xi-\xi_m\rVert\lVert\xi_n\rVert+\lvert((x-x_i)\xi_m|\xi_n)\rvert\\
&\leq4\epsilon+\lvert((x-x_i)\xi_m|\xi_n)\rvert,
\end{align*}
and the last term goes to $0$ for $i\to\infty$. This shows $x_i\to x$ in the wot. Again we have used that $x_i,x\in S$ and $\lVert\xi_k\rVert\leq1$ for all $k$.
