If $f_n,f\in C_0$ and for all $x_n,x$ with $x_n\to x$ we've got $f_n(x_n)\to f(x)$, then $\left\|f_n-f\right\|_\infty\to0$ Let $E$ be a locally compact Hausdorff space (if necessary, assume $E$ is a complete locally compact separable metric space) and $$C_0(E):=\left\{f\in C(E):\left\{|f|\ge\varepsilon\right\}\text{ is compact for all }\varepsilon>0\right\}.$$

How can we show that if $(f_n)_{n\in\mathbb N}\subseteq C_0(E)$ and $f\in C_0(E)$ with $$\left|f_n(x_n)-f(x)\right|\xrightarrow{n\to\infty}0\tag1$$ for all $(x_n)_{n\in\mathbb N}\subseteq E$ and $x\in E$ with $x_n\xrightarrow{n\to\infty}x$, then $$\left\|f_n-f\right\|_\infty\xrightarrow{n\to\infty}0\tag2,$$ where $\left\|\;\cdot\;\right\|_\infty$ denotes the supremum norm?

 A: The result is false. Take $E = \mathbb{R}$ and let $\phi$ be a bump function on $(-\varepsilon, \varepsilon)$. Let $f_n(x) = \phi(x - n)$. Then your first condition holds with $f = 0$ since $x_n$ is eventually close to $x$ and the support of $f_n$ is eventually far away from $x$ so for any sequence $x_n \to x$, $f_n(x_n)$ has a zero tail. 
However $\|f_n\|_\infty$ is constant. 

Your result is true if $E$ is a compact metric space. It is clear that $f_n \to f$ pointwise and so the sequence $f_n$ has at most one limit point in the $\sup$-norm. This means that by a standard subsequences argument, it is enough to check relative compactness. The rest of the argument looks long, but essentially just involves checking that your assumption implies the hypotheses of Arzela-Ascoli when $E$ is compact.
First, assume for a contradiction that the sequence $f_n$ is not equicontinuous. Then there is an $\varepsilon > 0$ such that for every $\delta > 0$, there is an $x$, a $y$ and an $n \in \mathbb{N}$ such that $d(x, y) < \delta$ and $$|f_{n}(x) - f_{n}(y)| \ge \varepsilon.$$
I claim we can arrange to have $n \to \infty$ as $\delta \to 0$ (along some sequence) instead of having $n$ be arbitrary. 
Set $\delta_0 = 1$. Then there are $n_0, x_0$ and $y_0$ such that $d(x_0,y_0) < \delta_0$ and $|f_{n_0}(x_0) - f_{n_0}(y_0)| \ge \varepsilon$.
Now suppose we have constructed $\delta_j, n_j, x_j$ and $y_j$ for $j < i$ with the properties that $j \mapsto n_j$ is strictly increasing, $\delta_j < \delta_{j-1} \wedge 2^{-j}$, $d(x_j,y_j)<\delta_j$ and $|f_{n_j}(x_j) - f_{n_j}(y_j)| \ge \varepsilon$. 
Since finite families of continuous functions on compact sets are easily seen to be equicontinuous, there is a $\delta_i < \delta_{i-1} \wedge 2^{-i}$ such that $d(x,y) < \delta_i$ implies that $|f_j(x) - f_j(y)| < \varepsilon$ for every $j \leq n_{i-1}$. Then since the sequence $(f_n)_{n \geq 1}$ is assumed to be not equicontinuous, there must exist an $n_i > n_{i-1}$ and an $x_i, y_i$ with $d(x_i,y_i) < \delta_i$ such that $|f_{n_i}(x_i) - f_{n_i}(y_i)| \ge \varepsilon$. 
Next, by compactness we can assume that there is an $x$ such that $x_i \to x$ and $y_i \to x$ by passing to a subsequence (the limit is the same since $d(x_i,y_i) \to 0$). This is where the argument first breaks without compactness.
All that remains is to bound
\begin{align}
\varepsilon \le |f_{n_i}(x_i) - f_{n_i}(y_i)| \le |f_{n_i}(x_i) - f(x)| + |f(x) - f_{n_i}(y_i)| \to 0
\end{align}
by our assumption, which gives a contradiction. Hence $(f_n)_{n \geq 1}$ is equicontinuous.
We are left to show that the sequence is uniformly bounded in the $\sup$-norm. To do this, by equicontinuity, fix a $\delta$ such that $d(x,y) < \delta$ implies that $|f_n(x) - f_n(y)| < 1$ for each $n$. 
By compactness of $E$, there is a finite set of points $\{x_i\}_{i \le K}$ that form a $\delta$-net for $E$. Then for any $n$ and any $y$, there is an $i \leq K$ such that $d(x_i,y) < \delta$ and so
\begin{align}
|f_n(y)| \leq & |f_n(y) - f_n(x_i)| + |f_n(x_i) - f(x_i)| + |f(x_i)| 
\\ \leq & 1 + \max_{i \leq K}|f_n(x_i) - f(x_i)| +  \max_{i \leq K}|f(x_i)|
\end{align}
uniformly in $y$. Then by our assumption, there is an $N$ such that for each $n \geq N$ and each $i$, $|f_n(x_i) - f(x_i)| < 1$. Putting this altogether implies that 
\begin{align}
{\|f_n\|}_\infty \leq 2 + \max_{i \leq K} |f(x_i)| + \max_{i < N} \|f_i\|_\infty < \infty
\end{align}
uniformly in $n$ so that $(f_n)_{n \geq 1}$ is uniformly bounded. Hence by Arzela-Ascoli it is relatively compact and hence must converge to $f$ in the $\sup$-norm.

It follows that your result also holds if $E$ is a locally compact, separable metric space and additionally $|f_n(x_n)| \to 0$ whenever $x_n \to \infty$ (in the sense that for every compact set $C \subset E$, $x_n \not \in C$ eventually). Indeed, in this case it is clear that if we take the Alexandroff compactification $E^* = E \cup \{\infty\}$ of $E$, then each $f_n$ can be identified with $f_n^* \in C(E^*)$ such that $f_n^*(\infty) = 0$, as can $f$. Further $E^*$ is a compact metric space and $E$ can be homeomorphically embedded in $E^*$.
The additional assumption on $f_n$ is exactly what is needed to mean your original assumption holds for $f_n^*$ and $f^*$ and hence by the above result, $\|f_n^* - f^*\|_{\infty,E^*} \to 0$. The result follows by noting that $\|f_n - f\|_{\infty,E} = \|f_n^* - f^*\|_{\infty,E^*}$. 
