How do I prove that the character equals to the psuedocharacter in compact Hausdroff spaces? I'm trying to prove, without success, that $\chi(x,X)=\psi(x,X)$ for all $x\in X$ with $X$ a compact Hausdorff space. In «Fundamentals of General Topology : Problems and Exercises» there's an excercise with a stronger result (that I can't prove either): 

$\chi(F,X)=\psi(F,X)$ for all $F\subseteq X$ closed with $X$ a compact Hausdorff space.

It's well known that $\psi(x,X)\leq\chi(x,X)$ for all $x\in X$ with $X$ a $T_1$-space due to the fact that every local base is a local pseudobase.
I've trying to prove that every local pseudobase is a local base but I realize that this is not true. How do I proceed?
 A: Let $\mathcal U$ be a family of open subsets of $X$ such that $|\mathcal U|=\psi(F,X)$ and $F=\bigcap\mathcal U$. Since the space $X$ is compact, it is normal. So for each set $U\in\mathcal U$ there exists an open set $V$ in $X$ with $$F\subset V(U)\subset \overline{V(U)}\subset U.$$ 
Then $$F\subset\bigcap \{V(U):U\in\mathcal U\}\subset \bigcap \{\overline{V(U)}:U\in\mathcal U\}\subset\bigcap \{U:U\in\mathcal U\}\subset F,$$
So all inclusions here are equalities. 
We claim that the family $\{V(U):U\in\mathcal U\}$ is a subbase at $F$. Indeed, let $W$ be an arbitrary open subset $X$ containing $F$. Then $\bigcap \{\overline{V(U)}\setminus W:U\in\mathcal U\}=\varnothing$. Since the space $X$ is compact,  the family $\{\overline{V(U)}\setminus W:U\in\mathcal U\}$ of its closed subsets is not centered. Therefore there exists a finite subfamily $\mathcal U’\subset\mathcal U$ such that $F\subset \bigcap \{\overline{V(U)}:U\in\mathcal U’\}\subset W$. Since $|\{V(U):U\in\mathcal U\}|=\psi(F,X)$, it generates a base at $F$ of size $\psi(F,X)$.
