Let $K$ be a compact set and $\{V_{1},...,V_{n}\}$ an open covering for $K$. We know that there exists a partition of unity subordinate to this covering, that is, $g_{1},...,g_{n}$ are such that $\text{supp}(g_{i})\subseteq V_{i}$, $0\leq g_{i}\leq 1$ is continuous, and $\displaystyle\sum_{i=1}^{n}g_{i}(x)=1$ for all $x\in K$. The author claims that one can assume without loss of generality that $g_{i}(x_{i})=1$ for some $x_{i}\in V_{i}$.
My question is, how this can be done? I try to let $h_{i}=\dfrac{g_{i}}{\|g_{i}\|}$, then surely there is some $x_{i}\in V_{i}$ such that $h_{i}(x_{i})=1$, but then it does not necessarily follow that $\displaystyle\sum_{i=1}^{n}h_{i}(x)=1$ for all $x\in K$.
I wonder if any trick is going on to create such a partition of unity.
Are you allowed to pass to a subcover? If so, by induction you can always pass to a subcover such that no $V_i$ is contained in the union of the $V_j$ for $j\ne i$. Thus, each $V_i$ has some point $x_i$ not contained in any other $V_j$, and therefore, if we have some partition of unity $(g_i)$ subordinate to this new covering, we know that $g_i(x_i)$ must be equal to $1$ for each $i$.