the problem says
Prove that a $T_{1}$ space $X$ is normal if and only if for each finite covering $\{U_{1},\dots,U_{n}\}$ of $X$ by open sets, there exist continuous functions $f_{1},\dots,f_{n}$ of $X$ into $I$ such that $f_{i} (x) = 0$ for $x \notin U_{i}$ and $\displaystyle \sum f_{i} (x) = 1$ for all $x \in X.$
*$I=[0,1]$
I'm applying Urysohn's lemma so I already have the part of $f_{i} (x) = 0$ for $x \notin U_{i},$ But the sum I can not see from where to get it.
Use that $X$ being normal implies that there are closed sets $F_1, \ldots, F_n$ with $F_i \subseteq U_i$ for all $i$ and $\cup_i^n F_i = X$ (the so-called shrinking lemma, which holds for finite open covers of normal spaces (even for point-finite infinite ones). Now apply Urysohn's lemma to get functions $g_i: X \to [0,1]$ with $g_i[F_i] = \{1\}$ and $g_i[X\setminus U_i] = \{0\}$, for all $i$.
Define $s(x) = \sum_{i=1}^n g_i(x)$. Note that $s(x) > 0$ for all $x$, and $s$ is clearly continuous.
Now define $f_i(x) = \frac{g_i(x)}{s(x)}$ for all $i$, and check that these are as required.
For the reverse direction, take $C,D$ disjoint and closed, and define $U = X\setminus C$, and $V= X\setminus D$. Then check that $U,V$ are open and cover $X$. Take $f_1, f_2$ as promised, then $f_1[C] = \{0\}$ and $f_2[D] = \{0\}$. The sum property implies that $f_2[C] = \{1\}$, and $f_1[D] = \{1\}$. Now $C \subseteq f_1^{-1}[[0,\frac{1}{2})]$ and $D \subseteq f_1^{-1}[(\frac{1}{2},1]]$ and these are disjoint open sets separating $C$ and $D$.