What is wrong with this fake proof that subgroup of a cyclic group is cyclic?

You are correct in proving that, for every $x\in G$, $\langle x\rangle\ne H$. But this doesn't imply that, for every $x\in G$, there is $h\in H$ such that $h\notin\langle x\rangle$.

For instance, this is false as soon as $H\ne G$ and $x$ is a generator of $G$.

No $s\in G$ such that $\langle s\rangle=H$ is clear.

But then you can't follow with for all $s\in G$ there's $h\in H$ such that $h\notin\langle s\rangle$ because $H$ could be a subset of $\langle s\rangle$ even though $s\notin G$.

Which, incidentally, is what happens if $s$ is a generator of $G$