We first need a sequence $(a_{k})_{k\in\mathbb{N}}$ in $K$ and a sequence $(U_{k})_{k\in\mathbb{N}}$ of disjoint open sets in $K$ with compact closure such that $U_{n}\cap \{a_{k} : k\in\mathbb{N} \} = \{a_{n}\}$ for all $n\in\mathbb{N}$. To establish the existence of such sequences, see here. Now you can proceed with your construction as follows. Let $n\in\mathbb{N}$. By Urysohn's lemma for locally compact Hausdorff spaces (see here) there is a continuous $f_{n}\colon K \to [0,1]$ such that $f_{n}(a_{n}) = 1$ and ${\rm supp} \, f_{n} \subseteq K\setminus U_{n}$. In particular, we have $f_{n}(a_{k}) = 0$ for all $k\in\mathbb{N}\setminus \{n\}$.
We now wish to show the map $\Phi \colon c_{0} \to C_{0}(K)$ given by
\begin{equation}
(\Phi (\alpha_{k})_{k\in\mathbb{N}})(x) := \sum_{k=1}^{\infty}\alpha_{k} f_{k}(x)
\end{equation}
is a well-defined linear isometry. Fix a sequence $(\alpha_{k})_{k\in\mathbb{N}} \in c_{0}$. For each $n\in\mathbb{N}$ define $g_{n} \colon K \to \mathbb{K}$ by $g_{n}(x) := \sum_{k=1}^{n} \alpha_{k} f_{k}(x)$. Since $U_{k}$ has compact closure for every $k\in\mathbb{N}$ it follows that $g_{n} \in C_{0}(K)$ for all $n\in\mathbb{N}$. Now let $m,n\in\mathbb{N}$ with $n < m$. Let $x\in K$. Using that $({\rm supp} \, f_{k} )_{k\in\mathbb{N}}$ is a sequence of disjoint sets and that $f_{k}$ takes values in $[0,1]$ for each $k\in\mathbb{N}$, we obtain
\begin{equation}
|g_{n}(x) - g_{m}(x)| = \Big|\sum_{k=n+1}^{m} \alpha_{k} f_{k}(x) \Big| \leq \sum_{k=n+1}^{m} |\alpha_{k}| f_{k}(x) \leq \sup\{|\alpha_{k}| : k\in \mathbb{N}\setminus \{1, \ldots , n\} \}
\end{equation}
and hence
\begin{equation}
\|g_{n} - g_{m}\|_{\infty} \leq \sup\{|\alpha_{k}| : k\in \mathbb{N}\setminus \{1, \ldots , n\} \}.
\end{equation}
As $(\alpha_{k})_{k\in\mathbb{N}} \in c_{0}$ this shows that $(g_{n})_{n\in\mathbb{N}}$ is a Cauchy sequence in $C_{0}(K)$. By completeness the sequence $(g_{n})_{n\in\mathbb{N}}$ converges uniformly to $g \in C_{0}(K)$ given by $g(x) := \sum_{k=1}^{\infty} \alpha_{k} f_{k}(x)$. Now let $x\in K$. Using again that $({\rm supp} \, f_{k} )_{k\in\mathbb{N}}$ is a sequence of disjoint sets and that $f_{k}$ takes values in $[0,1]$ for each $k\in\mathbb{N}$, we have
\begin{equation}
|g(x)| \leq \sum_{k=1}^{\infty} |\alpha_{k}| f_{k}(x) \leq \sup \{|\alpha_{k}| : k\in\mathbb{N} \}
\end{equation}
for every $x\in K$. Hence $\|g\|_{\infty} \leq \sup \{|\alpha_{k}| : k\in\mathbb{N} \}$. On the other hand, $g (a_{n}) = \alpha_{n}$ for every $n\in\mathbb{N}$ and it follows that $\sup \{|\alpha_{k}| : k\in\mathbb{N} \} \leq \|g\|_{\infty}$. Hence $\|g\|_{\infty} = \sup \{|\alpha_{k}| : k\in\mathbb{N} \}$. From what has been shown the map $\Phi$ is a well-defined isometry. Since $\Phi$ is clearly linear, we are done.
