Let $\{f_n\}$ be a sequence of continuous function on a metric space $E$ and $\lim_{n\to \infty} f_n(x_n)=f(x)$ for every $x_n\to x$ and $x\in E $ . 
Let $E$ be a metric space where every point of $E$ is an accumulation point. Let $\{f_n\}$ be a sequence of continuous function on $E$ and $$\lim_{n\to \infty} f_n(x_n)=f(x)$$ for every sequence of point $x_n \in E$ such that $x_n\to x$ and $x_n\neq x $ . Can we have $f_n \to f $ pointwisely or uniformly ?  

My attempt:
Let $E=(0,1]$ and $$f_n(x) = \begin{cases} n ,  0 < x < \frac1n \\
      \frac1x  ,  \frac{1}{n} \le x \le 1 \end{cases}$$
Then we can easily see $f_n(x)$ satisfied the hypothesis above and $f_n \to \frac1x$ , but converges not uniformly .  
My question :
$(1)$ Can we construct $\{f_n\}$ such that $\lim_{n\to \infty} f_n(x) \neq f(x)$ for some $x \in E$
$(2)$ If we assume $E$ is compact , can we show that $f_n \to f$ pointwisely or uniformly ? 
 A: As pointed out in the comment, under your hypothesis $(f_n(x))_{n\in\mathbb{N}}$ could diverge.
But if you add another condition that every point of $E$ is an accumulation point, we can say

  
*
  
*$f_n \to f $ pointwisely.
  

Proof $\,\,$   Suppose for contradiction, then we have $\exists$ $\varepsilon_0 >0$, $\forall$ $N$, $\exists$ $n\ge N$, such that $d\left(f_n(x),f(x)\right)\ge \varepsilon_0$ for some $x \in E$, and hence we can find a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ such that $d(f_{n_k}(x),f(x))\ge \varepsilon_0$.
Considering that $x$ is an accumulation point and that $f_n$ is continuous, we can find a sequence $(x_k)_{k\in\mathbb{N}} \to x$ such that $d(f_{n_k}(x_k),f(x))\ge \frac{\varepsilon_0}{2}$ , which contradicts the hypothesis $\lim_{n\to\infty} f_n(x_n) = f(x)$ for every $x_n \to x$ (because the subsequence of a convergent sequence still converges to its limit).


  
*$f$ is continuous.
  

Proof $\,\,$ Suppose $(x_k)$ converges to $x$ and let $\varepsilon>0$. From proposition 1, we can choose a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ such that $ d(f(x_k),f_{n_k}(x_k))<\varepsilon$ for every $k$. The hypothesis that $\lim_{n\to\infty} f_n(x_n) = f(x)$ for every $x_n \to x$ implies $\lim_{k\to\infty} f_{n_k}(x_k) = f(x)$. 
Note that
$
d(f(x_k),f(x))
\le d(f(x_k),f_{n_k}(x_k))+d(f_{n_k}(x_k),f(x))
$ thus there exists $N$ such that $\forall n > N$ $d(f(x_n),f(x)) < 2 \varepsilon$. Thus $f$ is continuous.


  
*If $E$ is compact, then $f_n \to f$ uniformly.
  

Proof $\,\,$ Suppose for contradiction, then we can find a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ and a sequence $(x_k)_{k\in\mathbb{N}}$ such that for every $k$, $d(f_{n_k}(x_k),f(x_k))\ge \varepsilon_0$ for some positive constant $\varepsilon_0$. 
Since $E$ is a compact metric space, then $E$ is limit point compact. Note that the points that appear in $(x_k)_{k\in\mathbb{N}}$ can't be finitely many, otherwise it would contradict proposition 1, and hence $(x_k)_{k\in\mathbb{N}}$ has an accumulation point, say $x$. Without loss of generality we suppose $x_k \to x$. Thus we have $\lim_{k\to\infty} f_{n_k}(x_k) = f(x)$.
Form proposition 2 we have $\lim_{k\to\infty} f(x_k) = f(x)$. Thus there exist $N$ such that $\forall n > N$ we have $d (f_{n_k}(x_k) ,f(x)) <\frac{\varepsilon_0}{2}$ and $d (f(x_k) ,f(x)) <\frac{\varepsilon_0}{2}$, which contradicts $d(f_{n_k}(x_k),f(x_k))\ge \varepsilon_0$.
