a theorem about converge pointwisely and uniformly Similar to the  Egoroff theorem, we can get the following theorem:

Theorem: Let $X$ be a locally compact Hausdorff
space(non-empty),and $\{f_n\}$ be a pointwise bounded sequence of
continuous functions defined on $X$,then $\{f_n\}$ are bounded
uniformly on an open subset of $X$

From this, I want to judge whether the theorem holds for pointwise convergent sequence:

Let  $X$ be a locally compact Hausdorff space(non-empty),and $\{f_n\}$
be a pointwise convergent  sequence of continuous functions defined on
$X$,then $\{f_n\}$ are convergent uniformly on an open subset of $X$.

Since a bounded sequence can contain a convergent subsequence, I guess the above theorem is also true, but I don't know where to start.
 A: Here is another example (perhaps in fact the same as David's). It is a bit too long for a comment.
Enumerate all non trivial closed sub-intervals of $[0,1]$ with rational endpoints as a sequence $(I_k)_{k\in\mathbb N}$. For each $k$, choose your favourite sequence of continuous functions converging pointwise to $0$ but not uniformly on $I_k$. More precisely, choose a sequence of continuous functions $(f_{n,k})_{n\in\mathbb N}$ on $[0,1]$ with $f_{n,k}\equiv 0$ outside $I_k$ and $0\leq f_{n,k}\leq 1$, such that $f_{n,k}\to 0$ pointwise as $n\to\infty$ but $\sup_{x\in I_k} f_{n,k}(x)=1$ for any $n,k$. 
Then define $f_n(x)=\sum_{k=1}^\infty 2^{-k}f_{n,k}(x)$. The $f_n$ are continuous (uniform convergence) with $0\leq f_n\leq 1$. It is not hard to check that $f_n\to 0$ pointwise (you have to interchange a limit and a $\Sigma$). On the other hand, for any $k, n$ you have $f_n(x)\geq 2^{-k} f_{k,n}(x)$; so $\sup_{x\in I_k} f_n(x)\geq 2^{-k}$ for each $k$ and all $n\in\mathbb N$, and hence no subsequence of $(f_n)$ can converge uniformly to $0$ on any $I_k$. Since any open subset of $[0,1]$ contains an $I_k$, it follows that no subsequence of $(f_n)$ converges uniformly on any open set.
