= Solution
For a <unital> <star-subalgebra> $A\subseteq B(H)$, the <Von Neumann double commutant theorem> states
$$
\boxed{\overline A^{\mathrm{SOT}}=\overline A^{\mathrm{WOT}}=A''.}
$$
Here $A'$ is the <commutant of an operator algebra>, consisting of operators commuting with every element of $A$. The double commutant is closed in either topology: multiplication by a fixed <bounded operator> is continuous in both. Therefore both closures of $A$ lie in $A''$.
For the converse, let $T\in A''$ and choose finitely many <vectors> $\xi_1,\ldots,\xi_n$. In $H^n$ put
$$
K=\overline{\{(a\xi_1,\ldots,a\xi_n):a\in A\}}.
$$
It is invariant under the diagonal action of $A$ and its adjoints, so its <orthogonal projection> $Q$ commutes with that diagonal action. Consequently each <matrix> entry $Q_{ij}$ belongs to $A'$. The diagonal operator with entry $T$ therefore commutes with $Q$. Since $I\in A$, the tuple $\xi=(\xi_1,\ldots,\xi_n)$ belongs to $K$, and hence so does $(T\xi_1,\ldots,T\xi_n)$. By the definition of $K$, one element $a\in A$ approximates this entire tuple arbitrarily closely. This is exactly approximation in every strong-operator neighborhood of $T$. It proves $A''\subseteq\overline A^{\mathrm{SOT}}$ and thus both asserted equalities. In particular, $M=A''$ is a <Von Neumann algebra>.
The <Kaplansky density theorem> strengthens the result by preserving <norm> bounds:
$$
\boxed{\overline{\{a\in A:\|a\|\leq1\}}^{\mathrm{SOT}}
=\{x\in M:\|x\|\leq1\}.}
$$
It also holds for the <self-adjoint> and positive unit balls; the full ball can in fact be approximated in the <strong-star topology>, meaning strong convergence of both operators and adjoints. We prove the bound rather than merely scaling the unbounded approximants furnished by the double commutant theorem.
First put $C=\overline A^{\|\cdot\|}$, a <unital> <C-star algebra>. Its <self-adjoint> part is weak-operator dense in $M_{\mathrm{sa}}$: symmetrize any weak approximating net. For a convex subset, strong and weak operator closures agree. To see this explicitly, separation from a strong <closure> supplies a continuous real functional on finitely many image <vectors>, hence one of the form $\operatorname{Re}\sum_j\langle a\xi_j,\eta_j\rangle$, which is also weak-operator continuous. The <Hahn-Banach separation theorem> would then separate the point from the weak <closure> as well. Thus every <self-adjoint> $x\in M$ is a strong limit of <self-adjoint> $a_i\in C$, initially without a common <norm> bound.
The needed functional-calculus <continuity> follows directly from <resolvents>. The <resolvent identity> gives
$$
(a_i-iI)^{-1}-(x-iI)^{-1}
=(a_i-iI)^{-1}(x-a_i)(x-iI)^{-1}.
$$
The first factor has <norm> at most one, so the right side tends strongly to zero. The same holds at $-i$. Both <resolvent> nets are uniformly bounded, so their products converge strongly. Their scalar functions generate a <self-adjoint> algebra separating points of $\mathbb R$ and vanishing at infinity. The <Stone-Weierstrass theorem>, applied to the one-point compactification, makes this algebra uniformly dense in $C_0(\mathbb R)$. Uniform approximation and the functional-calculus <norm> bound therefore show
$$
f(a_i)\longrightarrow f(x)\text{ strongly for every }f\in C_0(\mathbb R).
$$
This proves <strong continuity of functional calculus through resolvents> without an unjustified uniform bound on the original $a_i$.
If $x=x^*$ and $\|x\|\leq1$, choose a real compactly supported continuous $f$ equal to the identity on $[-1,1]$ and bounded in absolute value by one. Then $f(a_i)\in C$ are <self-adjoint> contractions and converge strongly to $f(x)=x$. For a positive contraction $x$, choose instead $0\leq f\leq1$ equal to the identity on $[0,1]$; this gives positive contractions. These prove the two restricted versions for $C$.
For an arbitrary contraction $x\in M$, apply the <self-adjoint> result in the <matrix algebra> $M_2(C)$ to
$$
X=\begin{pmatrix}0&x\\x^*&0\end{pmatrix},\qquad \|X\|=\|x\|\leq1.
$$
The strong <closure> of $M_2(C)$ is $M_2(M)$, since finitely many entries may be approximated simultaneously. The upper-right entries $c_i$ of the approximating <self-adjoint> contraction <matrices> belong to $C$, satisfy $\|c_i\|\leq1$, and converge strongly to $x$. The lower-left entries are $c_i^*$ and converge strongly to $x^*$. This proves the full strong-star assertion for $C$.
Finally transfer the bounds from $C$ to the possibly nonclosed algebra $A$. A contraction $c\in C$ has an approximant $b\in A$ with $\|b-c\|<\varepsilon$, so $b/(1+\varepsilon)$ is an $A$-contraction within $2\varepsilon$ of $c$. <Self-adjoint> approximants are obtained by symmetrizing first. For a positive contraction, approximate $c^{1/2}$ by $b\in A$ and use the positive contraction $b^*b/(1+\varepsilon)^2$, which tends in <norm> to $c$ as $\varepsilon\downarrow0$. Combining these <norm> approximations with each finite-vector strong approximation proves all versions for $A$. The reverse inclusions follow because the relevant <norm> bounds and the inequalities defining a <positive operator> are preserved by strong limits.
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