Every locally compact metric space can be given a compatible complete metric.
Suppose that $X$ is a locally compact metric space. Then $X$ is paracompact, so $X$ is a disjoint union of $\sigma$-compact locally compact spaces (There is a theorem proved in the book Topology by Dugundji that proves that every paracompact locally compact space is the disjoint union of $\sigma$-compact spaces. I prove this fact below.). Therefore we may without loss of generality assume that $X$ is $\sigma$-compact. Since we assume $X$ is $\sigma$-compact, there is a sequence of open sets $U_{n}$ such that $\overline{U_{n}}$ is compact and $\overline{U_{n}}\subseteq U_{n+1}$ for all $n$. From this sequence of open sets and Urysohn's lemma, there is a function $f:X\rightarrow[0,\infty)$ such that $\overline{U_{n}}\subseteq f^{-1}[0,n)\subseteq U_{n+1}$. Let $d$ be a metric on $X$, and define a new metric $d'$ on $X$ by letting $d'(x,y)=d(x,y)+|f(x)-f(y)|$. Clearly $(X,d')$ induces the original topology on $X$. I claim that $(X,d')$ is complete. Assume that $(x_{n})_{n}$ is a Cauchy sequence in $(X,d')$. Then the sequence $(x_{n})_{n}$ is bounded in $(X,d')$, so clearly the sequence $(f(x_{n}))_{n}$ is bounded as well. Therefore, there is some $N$ where $f(x_{n})<N$ for all $n$. In particular, since $f(x_{n})\subseteq f^{-1}[0,N)$, we have $x_{n}\in U_{N+1}\subseteq\overline{U_{N+1}}$ for all $n$. Since $\overline{U_{N+1}}$ is compact, and $x_{n}\in\overline{U_{N+1}}$ for all $n$, the sequence $(x_{n})_{n}$ has a convergent subsequence, so the sequence $(x_{n})_{n}$ itself must be convergent.
Added Later I will now prove that a paracompact locally compact space is a free union of $\sigma$-compact spaces since the proof is not too difficult.
Suppose that $X$ is locally compact and paracompact. Then let $\mathcal{U}$ be a cover of $X$ such that if $U\in\mathcal{U}$, then $\overline{U}$ is compact. Then let $\mathcal{V}$ be a locally finite open refinement of $\mathcal{U}$.
I claim that each $U\in\mathcal{V}$ intersects only finitely many other elements in $\mathcal{V}$. Since $\mathcal{V}$ is locally finite, there is an open cover $\mathcal{O}$ of $X$ where each $O\in\mathcal{O}$ intersects only finitely many elements of $\mathcal{V}$. If $U\in\mathcal{V}$, then since $\overline{U}$ is compact, there are $O_{1},...,O_{n}\in\mathcal{O}$ where $U\subseteq\overline{U}\subseteq O_{1}\cup...\cup O_{n}$. However, since the sets $O_{1},...,O_{n}$ each only intersect finitely many elements of $\mathcal{V}$, the union $O_{1}\cup...\cup O_{n}$ can only intersect finitely many elements of $\mathcal{V}$. In particular, the set $U$ only intersects finitely many elements of $\mathcal{V}$. Now make the set $\mathcal{V}$ into a graph where we put an edge between $U,V\in\mathcal{V}$ if and only if $U\cap V\neq\emptyset$. Then each $U\in\mathcal{V}$ has only finitely edges. Let $(\mathcal{E}_{i})_{i\in I}$ be the set of all connected components of the graph $\mathcal{V}$. Then each $\mathcal{E}_{i}$ is a countable subset and clearly $\{\bigcup\mathcal{E}_{i}|i\in I\}$ partitions $X$ into clopen subsets. Therefore since each $\mathcal{E}_{i}$ is countable and each $U\in\mathcal{V}$ has compact closure, the each union $\bigcup\mathcal{E}_{i}$ is $\sigma$-compact.
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For a reference: this paper has a reference [30] that has a proof. The author cites your result and refers to it. I don't have access to these papers, so I cannot verify exactly.