Suppose $X-A=\bigcup_{i=1}^{\infty}F_i$ is an open $G_\sigma$-set, i.e., $F_i$ are opens, in a normal space $X$. Then by Urysohn's lemma, for each i, there exists a continuous function $f_i:X\longrightarrow [0,1]$ such that
$$f_i(x)=0 \text{ for } x\in A \text{ and } f_i(x)=1 \text{ for } x\in F_i.$$ Now define $f(x)=\sum_{i=1}^{\infty}\frac{1}{2^i}f_i(x)$. My question is why $\{f_i\}$ uniformly convergent to $f(x)$? To specific, for every $\varepsilon>0$ there exists a $k$ such that $|f(x)-f_i(x)|<\varepsilon$ for every $x\in X$ and $i\geq k$.
It is simply not true. Use that $$\sup_{x\in X} \left\vert \frac{1}{2^i}f_i (x) \right\vert =\frac{1}{2^i}.$$ If $(f_i)_{i\geq 1}$ would converge to $f$, this would imply that $f=0$. What is true (use the triangular inequality) is that the following: $$ \lim_{N\rightarrow \infty} \left\Vert f - \sum_{k=1}^N \frac{1}{2^i}f_i \right\Vert_{\sup} = 0.$$ I.e. the partial sum converge uniformly to $f$.