I want to show that the language $H_\epsilon=\{x \mid M_x(\epsilon )\text{ halts } \}$ is undecidable, by reducing $H_0$ to that.
We have that $M_x$ is the Turing machine that is coded by $x$ and $H_0$ is the restricted halting problem, $H_0=\{x\mid M_x(x)\text{ halts } \}$.
Suppose that $H_\epsilon$ is decidable. Then there is Turing machine that computes this language for every input. That means that it is also computed with input $x$, that means that $H_0$ is then also decidable. Contradiction.
Is this the correct idea for the proof?