Let $X$ be a locally compact metric space. Can we choose a sequence $H_i$ of compact sets such that $H_i \subset \operatorname{int}\left(H_{i+1}\right)$ for all $i \geq 1$ and $X =\cup_{i=1}^{\infty} H_i$.
Ignore the counterexample : uncountable set with discrete metric. Please, provide me some other examples if known.
On
In general it is impossible. A space $X$ which is the countable union of compact subsets is called $\sigma$-compact. It is well-known that if $X$ is locally compact, then $X$ is $\sigma$-compact if and and only it is representable as in your question.
So your question is: Is every locally compact space which is not discrete a $\sigma$-compact space?
Consider an uncountable product $P = \prod_{\alpha \in A} I$ of unit intervals $I = [0,1]$ and $x \in P$. Then $X = P \setminus \{ x \}$ is locally compact. If it were the countable union of compact subsets, then $\{ x \}$ would be the intersection of countably many open neighborhoods. A basis of open neighborhoods is given by the family of sets $\bigcap_{i=1}^n p_{\alpha_i} ^{-1}(U_i)$, where $p_\alpha : P \to I$ is the projection onto the coordinate $\alpha$, $U_i \subset I$ an open neighborhood of $x_i = p_{\alpha_i}(x)$ and $n \in \mathbb{N}$ is arbitrary. But it is easy to see that any countable intersection of such sets is still an uncountable set since only countably many coordinates are subject to restrictions.
This holds (for metric spaces) iff $X$ is separable as well:
If $X$ is locally compact metric and we have the $H_i$ as promised, $X$ is $\sigma$-compact, hence Lindelöf and separable and second countable (as these are equivalent in all metric spaces).
If $X$ is locally compact metric and separable, we can reduce the base of open sets with compact closure to a countable base $\{B_n: n \in \omega\}$ with all $\overline{B_n}$ compact. Then set $H_0 = B_0$ and cover $\overline{B_0}$ by finitely many new $B_n$, whose union we then define to be $H_1$ etc. continuing by recursion.