Is my proof of $C_G(H) \le N_G(H)$ correct? Let $x\in C_G(H)$. This means $xh = hx$ for all $h \in H$. Then $xH = Hx$ (This is that part I'm not so sure about). Hence, $x \in N_G(H)$, so that we have $C_G(H) \le N_G(H)$.

Yes that's completely correct. To explicitly see the part you're unsure about, since $$xH := \{xh:h\in H\}$$
and $x\in C_G(H)$ so that $xh = hx$ for all $h\in H$, it is indeed the case that
$$xH = \{xh:h\in H\}=\{hx:h\in H\} =Hx.$$
