I need the following result for a problem that I am solving. I would love to have a critique for my proof.
A net $(A_{\lambda})$ in $M_n(B(H))=M_n(H^n)$ can be written as a matrix $(a_{{ij},\lambda})$.
therefore the net $A_\lambda \rightarrow 0$ iff $a_{{ij}} \rightarrow 0$ for all $i,j$ by identifying $M_n(B(H))$ with $M_n(H^{n})$
First of all, $M_n(B(H))$ is not identified with $M_n(H^n)$, this doesn't really make any sense. There is a $*$-isomorphism (isometric) $M_n(B(H))\cong B(H^n)$ and the correspondence is $(a_{i,j})_{i,j}\longleftrightarrow A$ where $A$ is the operator given by $$A(h_1,\dots,h_n)=\big{(}\sum_{j=1}^na_{1j}h_j,\dots,\sum_{j=1}^na_{nj}h_j\big{)}.$$
The WOT makes sense on $M_n(B(H))$ only through this isomorphism. Therefore a net of matrices $((a_{i,j}^\lambda)_{i,j})_{\lambda\in\Lambda}$ will converge WOT-wise in $M_n(B(H))$ to the $0$ matrix if and only if the net $(A_\lambda)_{\lambda\in\Lambda}\subset B(H)$ of the corresponding operators converges WOT-wise to the corresponding operator of the $0$ matrix, which is the $0$ operator.
Therefore, $((a_{i,j}^\lambda)_{i,j})\to 0$ in WOT if and only if we have that for any $(h_1,\dots,h_n),(h_1',\dots,h_n')\in H^n$ it is $$\sum_{i=1}^n\sum_{j=1}^n\langle a_{ij}^\lambda h_j,h_i'\rangle\to0.$$ Conclusion: If for all $i,j$ we have $a_{ij}^\lambda\to 0$ in WOT of $B(H)$, then yes, the matrix net converges WOT-wise to $0$ in $M_n(B(H))$.
Is the converse true? Sure: suppose that $((a_{i,j}^\lambda)_{i,j})_{\lambda})$ converges to $0$ in WOT of $M_n(B(H))$, so we have that
$$\sum_{i,j=1}^n\langle a_{ij}^\lambda h_j,h'_i\rangle\xrightarrow{\lambda\in\Lambda}0 $$ for all $(h_1,\dots,h_n)$ and $(h_1',\dots,h_n')$ in $H^n$. Fix indexes $k$ and $l$ and arbitrary vectors $x,y$ in $H$ and consider $(h_1,\dots,h_n)$ to be the vector which has $0$ everywhere except the $l$ slot, where it has $x$ and likewise let $(h_1',\dots,h_n')$ be the vector with 0 everywhere except the $k$ slot where it has $y$. By the above relation applied to those vectors, we have that $\langle a_{kl}^\lambda x,y\rangle\to0$, showing that $(a_{kl}^\lambda)_{\lambda}$ converges to $0$ in the WOT of $B(H)$.