Every paracompact Hausdorff space is regular

For every $x\in F$, Hausdorffness gives disjoint open sets $U,V$ with $x\in U$ and $p\in V$; hence $p\mathrm{\notin }\bar{U}$, since $X\setminus V$ is closed and contains $U$. Thus the family of all open $U$ with $U\cap F\mathrm{\ne }\mathrm{\varnothing }$ and $p\mathrm{\notin }\bar{U}$, together with $X\setminus F$, is an open cover $\mathcal{U}$ of $X$.

Take a locally finite open cover $\mathcal{W}$ refining $\mathcal{U}$, and put $H:=\bigcup \{W\in \mathcal{W}:W\cap F\mathrm{\ne }\mathrm{\varnothing }\}$.

The set $H$ is open and contains $F$: a member of $\mathcal{W}$ containing a point of $F$ cannot refine $X\setminus F$, so it occurs in the defining union.

Every $W$ occurring in $H$ lies in an eligible $U$ of step 1.1, so $p\mathrm{\notin }\bar{W}$; local finiteness and [L1] give $\bar{H}=\bigcup \bar{W}$, whence $p\mathrm{\notin }\bar{H}$.

The open sets $X\setminus \bar{H}$ and $H$ contain $p$ and $F$ respectively and are disjoint. By [F3], $X$ is regular.