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Every locally compact metric space can be given a compatible complete uniformity. 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 Dugundji that proves that every paracompact locally compact space is the countable union of $\sigma$-compact spaces). 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.