[proofplan]
The inclusion $\overline{A}^{\mathrm{SOT}}\subseteq \overline{A}^{\mathrm{WOT}}$ follows because strong operator convergence implies weak operator convergence. For the reverse inclusion, we take $T$ in the weak operator closure and prove that every basic strong neighbourhood of $T$ meets $A$. The key finite-dimensional packaging is to put the vectors $T\xi_1,\dots,T\xi_n$ into $H^{\oplus n}$ and prove that this vector lies in the closed subspace generated by $(a\xi_1,\dots,a\xi_n)$ with $a\in A$. Orthogonality against the complement is detected by weak-operator continuous matrix coefficient functionals, so weak closure forces the needed orthogonality.
[/proofplan]
[step:Use strong convergence to obtain weak convergence]
By [citetheorem:9260], if a net $(S_i)_{i\in I}$ in $\mathcal{L}(H)$ converges strongly to $S\in\mathcal{L}(H)$, then it converges to $S$ in the weak operator topology. Therefore every strong operator [limit point](/page/Limit%20Point) of $A$ is a weak operator limit point of $A$, and hence
\begin{align*}
\overline{A}^{\mathrm{SOT}}\subseteq \overline{A}^{\mathrm{WOT}}.
\end{align*}
[/step]
[step:Reduce strong approximation to a closed subspace in a finite direct sum]
Let $T\in \overline{A}^{\mathrm{WOT}}$. To prove $T\in\overline{A}^{\mathrm{SOT}}$, fix an integer $n\in\mathbb{N}$, vectors $\xi_1,\dots,\xi_n\in H$, and a tolerance $\varepsilon>0$. It is enough to find $a\in A$ such that
\begin{align*}
\|a\xi_r-T\xi_r\|_H<\varepsilon
\end{align*}
for every $r\in\{1,\dots,n\}$, because a basic strong operator neighbourhood of $T$ is specified by finitely many vectors together with finitely many tolerances, and replacing those tolerances by a common smaller $\varepsilon$ still gives a neighbourhood of the same form.
Let $H^{\oplus n}$ denote the [Hilbert space](/page/Hilbert%20Space) direct sum of $n$ copies of $H$, with [inner product](/page/Inner%20Product)
\begin{align*}
((\eta_1,\dots,\eta_n),(\zeta_1,\dots,\zeta_n))_{H^{\oplus n}}=\sum_{r=1}^{n}(\eta_r,\zeta_r)_H.
\end{align*}
Define the subset $E\subseteq H^{\oplus n}$ by
\begin{align*}
E=\{(a\xi_1,\dots,a\xi_n):a\in A\}.
\end{align*}
Since $A$ is a linear subspace of $\mathcal{L}(H)$, $E$ is a linear subspace of $H^{\oplus n}$. Let
\begin{align*}
K=\overline{E}^{\|\cdot\|_{H^{\oplus n}}}.
\end{align*}
We will prove that
\begin{align*}
(T\xi_1,\dots,T\xi_n)\in K.
\end{align*}
[guided]
We want strong approximation of $T$ by elements of $A$ on the finite set $\{\xi_1,\dots,\xi_n\}$. The strong operator topology is exactly the topology of pointwise norm convergence on $H$, so a basic strong neighbourhood of $T$ asks for an operator $a\in A$ with $a\xi_r$ close to $T\xi_r$ for finitely many prescribed vectors.
To package all these inequalities at once, we work in the Hilbert space $H^{\oplus n}$. This space consists of tuples $(\eta_1,\dots,\eta_n)$ with $\eta_r\in H$, and its inner product is
\begin{align*}
((\eta_1,\dots,\eta_n),(\zeta_1,\dots,\zeta_n))_{H^{\oplus n}}=\sum_{r=1}^{n}(\eta_r,\zeta_r)_H.
\end{align*}
For each $a\in A$, the tuple $(a\xi_1,\dots,a\xi_n)$ records how $a$ acts on the chosen finite set of vectors. Define
\begin{align*}
E=\{(a\xi_1,\dots,a\xi_n):a\in A\}\subseteq H^{\oplus n}.
\end{align*}
Because $A$ is closed under addition and scalar multiplication, $E$ is a linear subspace of $H^{\oplus n}$. Let
\begin{align*}
K=\overline{E}^{\|\cdot\|_{H^{\oplus n}}}
\end{align*}
be its norm closure.
The target tuple is $(T\xi_1,\dots,T\xi_n)$. If we prove
\begin{align*}
(T\xi_1,\dots,T\xi_n)\in K,
\end{align*}
then by the definition of norm closure there exists $a\in A$ with
\begin{align*}
\|(a\xi_1,\dots,a\xi_n)-(T\xi_1,\dots,T\xi_n)\|_{H^{\oplus n}}<\varepsilon.
\end{align*}
Since the direct-sum norm dominates each coordinate norm, this gives
\begin{align*}
\|a\xi_r-T\xi_r\|_H<\varepsilon
\end{align*}
for every $r\in\{1,\dots,n\}$. Thus membership of the target tuple in $K$ is precisely the finite-vector strong approximation we need.
[/guided]
[/step]
[step:Annihilate the orthogonal complement using weak operator continuity]
Let $\zeta=(\zeta_1,\dots,\zeta_n)\in K^\perp$. Define a linear functional
\begin{align*}
\Phi:\mathcal{L}(H)\to\mathbb{C}
\end{align*}
by
\begin{align*}
\Phi(S)=\sum_{r=1}^{n}(S\xi_r,\zeta_r)_H.
\end{align*}
Each map $S\mapsto (S\xi_r,\zeta_r)_H$ is one of the defining matrix coefficient maps for the weak operator topology, so $\Phi$ is weak-operator continuous as a finite linear combination of weak-operator continuous maps.
For every $a\in A$, the vector $(a\xi_1,\dots,a\xi_n)$ belongs to $E\subseteq K$. Since $\zeta\in K^\perp$, we have
\begin{align*}
0=((a\xi_1,\dots,a\xi_n),\zeta)_{H^{\oplus n}}=\sum_{r=1}^{n}(a\xi_r,\zeta_r)_H=\Phi(a).
\end{align*}
Thus $\Phi$ vanishes on $A$. Since $T\in\overline{A}^{\mathrm{WOT}}$ and $\Phi$ is weak-operator continuous, $\Phi$ also vanishes at $T$:
\begin{align*}
\Phi(T)=0.
\end{align*}
Equivalently,
\begin{align*}
((T\xi_1,\dots,T\xi_n),\zeta)_{H^{\oplus n}}=0.
\end{align*}
Therefore $(T\xi_1,\dots,T\xi_n)$ is orthogonal to every vector in $K^\perp$.
[/step]
[step:Convert orthogonality into membership in the closed subspace]
Since $K$ is a closed linear subspace of the Hilbert space $H^{\oplus n}$, the Hilbert space orthogonal complement identity gives
\begin{align*}
(K^\perp)^\perp=K.
\end{align*}
The previous step proved that $(T\xi_1,\dots,T\xi_n)\in (K^\perp)^\perp$. Hence
\begin{align*}
(T\xi_1,\dots,T\xi_n)\in K.
\end{align*}
By the definition of $K$ as the norm closure of $E$, there exists $a\in A$ such that
\begin{align*}
\|(a\xi_1,\dots,a\xi_n)-(T\xi_1,\dots,T\xi_n)\|_{H^{\oplus n}}<\varepsilon.
\end{align*}
Consequently, for each $r\in\{1,\dots,n\}$,
\begin{align*}
\|a\xi_r-T\xi_r\|_H<\varepsilon.
\end{align*}
[/step]
[step:Conclude equality of the strong and weak operator closures]
The preceding step shows that every basic strong operator neighbourhood of $T$ intersects $A$. Therefore $T\in\overline{A}^{\mathrm{SOT}}$. Since $T\in\overline{A}^{\mathrm{WOT}}$ was arbitrary, we have
\begin{align*}
\overline{A}^{\mathrm{WOT}}\subseteq \overline{A}^{\mathrm{SOT}}.
\end{align*}
Together with the reverse inclusion proved at the beginning, this gives
\begin{align*}
\overline{A}^{\mathrm{SOT}}=\overline{A}^{\mathrm{WOT}}.
\end{align*}
[/step]
