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 $p\notin\overline U$, since $X\setminus V$ is closed and contains $U$. Thus the family of all open $U$ with $U\cap F\ne\varnothing$ and $p\notin\overline U$, 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 $H:=\bigcup\{W\in\mathcal W:W\cap F\ne\varnothing\}$.

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 $p\notin\overline W$; local finiteness and [L1] give $\overline H=\bigcup\overline W$, whence $p\notin\overline H$.

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