I write $B$ to be the unit ball (closed) in $B(H)$ i.e. $B=\{T\in B(H)|\ \lVert T\rVert\le 1\}$
I define $D_{h,k}:=\{x\in\Bbb{C}|\ |x|\le 1\}$ for $h,k\in \overline{B}_H$ where $\overline{B}_H=\{h\in H|\ \lVert h\rVert\le 1\}$. $D_{h,k}$ is compact subset of $\Bbb{C}$.
Then I define the product $$\displaystyle{D:=\prod\limits_{(h,k)\in\overline{B}_H^2}}D_{h,k}$$
Then $D$ is compact by Tychonoff.
Define $\tau:B\to D$ by $$(\tau(T))_{(h,k)}=\langle Th,k\rangle$$
I am able to show that $\tau$ is a homeomorphim on its image ($B$ is equipped with the weak operator topology).
But to show $B$ is compact I have to show $\text{Ran}(\tau)$ is closed.
Let $f_\alpha\in \text{Ran}(\tau)$ be a net such that $f_\alpha\to f\in D\implies f_\alpha(h,k)\to f(h,k)\ \forall \lVert h\rVert,\lVert k\rVert\le 1$.
As $f_\alpha\in \text{Ran}(\tau)$, $f_\alpha(h,k)=\langle T_a h,k\rangle \ \forall \lVert h\rVert,\lVert k\rVert\le 1$ for some $T_\alpha\in B$.
That is we have $f(h,k)=\lim \langle T_\alpha h,k\rangle\in D_{h,k}\ \forall \lVert h\rVert,\lVert k\rVert\le 1$
Now, it's easy to show that $\langle T_\alpha h,k\rangle$ converges for all $h,k\in H$.
Now my target is to define $T\in B$ such that $f(h,k)=\langle T h,k\rangle\ \forall \lVert h\rVert,\lVert k\rVert\le 1$
My guess is to define $$T(h)=\lim T_\alpha (h)$$
But before defining this I need to verify that $\{T_\alpha h\}$ converges in $H$, that I'm unable to prove. Can anyone help me with the proof of this step?
Thanks for help in advance.