I think the following argument works under your hypothesis:

Consider $\mathcal{B}=\{B_n\}$ a countable basis of the topology of $E$ such that $\overline{B_n}$ is compact for any $n$ (this exists since $E$ is a separable metric space, thus, it has a countable basis and then a basis like this is constructed using local compactness).

Now, start with $K_0= \overline{B_0}$. Now, given $K_n$, define $K_{n+1}$ by the union of $\overline{B_{n+1}}$ with the closure of a finite subcovering of $K_n$ so, it is compact (being finite union of compact sets) and $K_n \subset int(K_{n+1})$.

We get that $\bigcup K_n =E$ since it contains $\bigcup B_n=E$.