If $f(H) = H$, we're done. Otherwise, let $f(H) = H_{1}\subset H$. Now, if $f(H_{1}) = H_{1}$, we're done. Otherwise, $f(H_{1}) = H_{2}\subset H_{1}= f(H)$. And so on. So for each $i$, either $f(H_{i}) = H_{i}$ or $H_{i+1}= f(H_{i}) \subset H_{i}$. If there is no instance of the first, we have an infinite sequence of decreasing subsets, which is impossible.

This is a particular case of the following result due to Tarski and Knaster: Let $\mathcal L$ be a complete lattice, and $f: \mathcal L\to\mathcal L$ an increasing map. Then, the set of fixed points of $f$ is non-empty (and in fact forms a complete lattice).