Is this proof of $\liminf E_k \subset \limsup E_k $ correct?

It’s not quite right: you’ve actually assumed that $x\in\limsup_kE_k$, so the argument is circular. Recall that

$$\liminf_{k\in\Bbb N}E_k=\bigcup_{n\ge 0}\bigcap_{k\ge n}E_k\;;$$

this means that if $x\in\liminf_{k\in\Bbb N}E_k$, then there is some $n_0\in\Bbb N$ such that $x\in\bigcap\limits_{k\ge n_0}E_k$.

Since $$\limsup_{k\in\Bbb N}E_k=\bigcap_{n\ge 0}\bigcup_{k\ge n}E_k\;,$$ you have to use this somehow to show that $x\in\bigcup\limits_{k\ge n}E_k$ for every $n\in\Bbb N$. That’s not actually very hard: we know that $x\in E_k$ for each $k\ge n_0$, so if we just take $k=\max\{n,n_0\}$ then ... ?