Sequence of continuous functions converges uniformly. Does it imply the limit function is continuous?
For $\varepsilon>0$, there is $n$ such that $|f(z)-f_n(z)|<\dfrac{\varepsilon}{3}$ for all $z$, take one such $n$, and as $f_n$ are continuous at $a$, for the same $\varepsilon$ there is $\delta>0$ s.t. $|x-a|<\delta\,\Rightarrow |f_n(x)-f_n(a)|<\dfrac{\varepsilon}{3}$. So, we have
$|f(x)-f(a)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|<\varepsilon$.