[proofplan]
This is a direct corollary of the [WOT Compactness of the Unit Ball](/theorems/1244). Every Hilbert space is reflexive (via the [Riesz Representation Theorem](/theorems/218)), so the general criterion — that $\overline{B}_{\mathcal{L}(X,Y)}$ is WOT-compact if and only if $Y$ is reflexive — applies with $X = H$ and $Y = K$.
[/proofplan]
[step:Verify that every Hilbert space is reflexive]
The [Riesz Representation Theorem](/theorems/218) establishes that for any Hilbert space $K$, the map
\begin{align*}
\Phi: K &\to K^* \\
y &\mapsto (\cdot, y)_K
\end{align*}
is an isometric isomorphism (conjugate-linear in the complex case, linear in the real case). In particular, $K^*$ is itself a Hilbert space under the transported inner product, and applying the Riesz Representation Theorem to $K^*$ gives $(K^*)^* \cong K^*$. Composing the canonical embedding $J: K \to K^{**}$ with the identification $K^{**} \cong (K^*)^* \cong K^*$, one verifies that $J$ is surjective: for every $\xi \in K^{**}$, there exists $y \in K$ with $J(y) = \xi$. Therefore $K$ is reflexive.
More concretely: the canonical embedding $J: K \to K^{**}$, $J(y)(f) = f(y)$, is always isometric by the [Hahn-Banach Theorem](/theorems/879). To show surjectivity, take $\xi \in K^{**}$. The composition $\xi \circ \Phi^{-1}: K \to \mathbb{R}$ is a bounded linear functional on $K$ (where $\Phi^{-1}: K^* \to K$ is the inverse of the Riesz isomorphism). By the [Riesz Representation Theorem](/theorems/218) applied to $K$, there exists $z \in K$ with $\xi(\Phi^{-1}(w)) = (w, z)_K$ for all $w \in K$... this unwinds to the standard identification. The conclusion is that $J$ is surjective, so $K$ is reflexive.
[guided]
Every Hilbert space is reflexive, and this is a fundamental consequence of the [Riesz Representation Theorem](/theorems/218). The argument proceeds in two stages.
The Riesz Representation Theorem states that for a Hilbert space $K$, every bounded linear functional $f \in K^*$ has the form $f = (\cdot, y)_K$ for a unique $y \in K$, with $\|f\|_{K^*} = \|y\|_K$. This gives the isometric isomorphism $\Phi: K \to K^*$, $y \mapsto (\cdot, y)_K$.
To show reflexivity, we must prove the canonical embedding $J: K \to K^{**}$, defined by $J(y)(f) = f(y)$ for $f \in K^*$, is surjective. Let $\xi \in K^{**}$ be arbitrary. Consider the map $g: K \to \mathbb{R}$ defined by $g(y) = \xi(\Phi(y))$. Since $\Phi$ is a bounded linear map and $\xi$ is a bounded linear functional, $g$ is a bounded linear functional on $K$: $g \in K^*$. By the Riesz Representation Theorem, there exists $z \in K$ with $g(y) = (y, z)_K$ for all $y$. Now for any $f \in K^*$, write $f = \Phi(y_f)$ for a unique $y_f \in K$. Then:
\begin{align*}
\xi(f) = \xi(\Phi(y_f)) = g(y_f) = (y_f, z)_K = \Phi(z)(y_f) = f(z) = J(z)(f).
\end{align*}
Since this holds for every $f \in K^*$, $\xi = J(z)$. Since $\xi$ was arbitrary, $J$ is surjective, and $K$ is reflexive.
[/guided]
[/step]
[step:Apply the WOT compactness criterion to conclude]
Since $K$ is reflexive, the [WOT Compactness of the Unit Ball](/theorems/1244) (applied with $X = H$ and $Y = K$, both Banach spaces) states that $\overline{B}_{\mathcal{L}(H, K)} = \{T \in \mathcal{L}(H, K) : \|T\| \le 1\}$ is compact in the WOT.
[/step]
