An unusual finite "partition of unity" on a compact Hausdorff space $X$ My question is very similar to Finite Number of Partitions of Unity in a Compact Hausdorff Space, but my problem doesn't involve compact supports, so I think this question is different: (I wasn't able to adapt its answer to my problem)

Let $X$ be a compact $T_{2}$ topological space, and let $X = \bigcup_{\lambda \in \Lambda} U_{\lambda}$ be an open covering of $X$. Prove that there exists an $n \in \mathbb{N}$, and continuous functions $f_{1},...,f_{n}:X \to [0,1]$ such that:

*

*$(\forall i \in \{1,...,n\})(\exists \lambda_{i} \in \Lambda) f_{i}|_{U_{\lambda_{i}}^{c}} \equiv 0$


*$(\forall x \in X) \sum_{i=1}^{n} f_{i}(x)=1.$

So first as first, obviously, I take a finite subcover $U_{\lambda_{1}},...,U_{\lambda_{n+1}}$ , and I apply Urysohn's lemma on the complements of the sets $V_{1} = \bigcup_{i=1}^{n+1} U_{\lambda_{i}}$, $V_{2} = U_{n+1} $ (if $n+1=1$, the matter is trivial), and I get functions $f$, $1-f = g$, where $f$ is taken from Urysohn's lemma, so $f(V_{1}^{c}) = \{0\}$, $g(V_{2}^{c}) = \{0\}$. Now, the subspace $V_{2}^{c}$ is also compact and $T_{2}$, since it's closed, so now I want to repeat this process, but I have no induction hypthesis.
So now I get the following idea, similar to the answer in the linked question: I want to prove that for every $n$, and every $X = \bigcup_{i=1}^{n} U_{i}$, where $U_{i}$ are open, I can construct functions $f_{1},...,f_{n}$ satisfying 1. and 2.
For $n=2$, I use Urysohn's lemma as described.
For $n \implies n+1$, I proceed as described as well , and then I use the induction hypothesis on $V_{2}^{c}$, and the cover $V_{2}^{c} = \bigcup_{i=1}^{n} (U_{i} \cap V_{2}^{c})$ getting $f_{1}, f_{2}, ..., f_{n}: V_{2}^{c} \to [0,1]$ such that $f_{i}|_{V_{2}^{c} \setminus U_{i}} \equiv 0$ and 2. for $X = V_{2}^{c}$. However, I don't know how to extend $f_{1},..., f_{n}$ to $X$. Obviously, I want $f_{i}\equiv 0$ on $(X \setminus U_{i}) \setminus (V_{2}^{c} \setminus U_{i})$, but what do I do on $U_{i} \cap V_{2}$? And if I do figure out how to define $f_{i}$ on $U_{i} \cap V_{2}$, how do I prove that that extension is continuous?
 A: 
Lemma (shrinking lemma): Let $X$ be a normal space. if $U_1, \ldots, U_n$ is a finite open cover of $X$, there is a closed cover $F_1, \ldots, F_n$ of $X$ such that $F_i \subseteq U_i$ for all $i\in \{1,\ldots,n\}$.

The proof can be done by induction, where the $n=2$ case is the crux:
Suppose $U_1 \cup U_2 = X$. Then $C_i = X\setminus U_i$ are closed and disjoint (two sets form a cover iff their complements are disjoint), so by normality we find disjoint open sets $W_i$ with $C_i \subseteq W_i$ ($=1,2$) and then we can define $F_i = X\setminus W_i$, and see it is as required: $$F_i = X \setminus W_i \subseteq X\setminus C_i = U_i$$ where we use that $\setminus$ reverses inclusions. 
Now suppose (induction hypothesis) that the lemma holds for all normal spaces and open covers of size $n \ge 2$, and we are given an open cover $U_1,\ldots, U_n, U_{n+1}$ of size $n+1$.
Define $U = \bigcup_{i=1}^n U_i$ and apply the $n=2$ case to the open cover $\{U, U_{n+1}\}$ to get a closed cover $\{F, F_{n+1}\}$ of $X$ with $F \subseteq U$ and $F_{n+1} \subseteq U_{n+1}$. Now, $F$ is also normal as a closed subspace of $X$, and it has a (relatively open) cover $U_1 \cap F, \ldots, U_n \cap F$ of size $n$, so we apply the induction hypothesis to get $F_1,\ldots F_n$, a closed cover of $F$ with $F_i \subseteq U_i \cap F$ for all $i=1,\ldots n$. As $F$ is closed and all $F_i$ are closed in $F$, all $F_i$ are closed in $X$ and now clearly the $F_1,\ldots F_n, F_{n+1}$ form a closed cover of $X$ with $F_i \subseteq U_i$ for all $i$. This finishes the induction step.
As your compact $X$ is normal (this is all we need) starting with a finite subcover $U_1, \ldots U_n$ we get the closed cover $F_i$ as in the lemma, and we apply Urysohn to get continuous functions 
$g_i : X \to [0,1]$ such that $g_i[F_i] = \{1\}$ and $g_i[X\setminus U_i] = \{0\}$, $i=1,\ldots, n$. 
Define $s(x)  = \sum_{i=1}^n g_i: X \to \mathbb{R}$ which is continuous as a finite sum of continuous functions. As the $F_i$ form a cover, for every $x \in X$, $x \in F_i$ for some $i$ and so $s(x) \ge g_i(x) = 1 > 0$, and hence $s$ is a strictly positive function. So defining $f_i(x) = \frac{g_i}{s(x)}$, $i=1,\ldots, n$, all $f_i$ are continuous (as division $(x,y) \to \frac{x}{y}$ is continuous on $\mathbb{R} \times (0,\infty)$) and the scaling now ensures that $\sum_{i=1}^n f_i(x) = 1$ for all $x$. $g_i$ vanishes outside $U_i$ and so does $f_i$ and so the $f_i$ are as required.
So we just use normality, essentially. 
Compactness is to reduce it to the finite case, which is easy to handle.
It turns out we can do this for all paracompact Hausdorff spaces and we need locally finite refinements. The shrinking lemma can be generalised to point-finite covers as well. 
