In the book Gaussian Measures by Bogachev, Theorem 2.3.1 he shows that if $\mu$ is a Gaussian measure on a separable Hilbert space, then its characteristic functional is given by $$\hat\mu(x)=e^{-i(a,x)-\frac{(Kx,x)}{2}}$$ where $a \in X$ is a vector and $K$ is a symmetric non negative nuclear operator. To show that $K$ is compact, he says
If $x_n\to 0$ in the weak topology, then $\hat\mu(x)\to 1$ by Lebesgue's theorem. Hence $\|\sqrt{K}x_n\|\to 0$
I don't see how this shows $K$ is compact.
This is a consequence of the Banach-Alaoglu theorem.
It tells us that on a separable Hilbert (or just reflexive) space the unit ball with the weak topology is metrizable and compact.
Observe that the quoted line implies that the operator $L:=\sqrt{K}$ restricted to the closed unit ball $B_X$ is weak-to-norm continuous as $B_X$ is metrizable. But then $L(B_X)$ is norm compact as $B_X$ is weakly compact. Hence, $L$ is compact and $K=L^2$ is compact as well.