It's unclear to me why would you start with a condition that is not given in the question. However yes, if $X$ was regular then $Y$ is regular . This will allow you to conclude Hausdorffness if you go by Munkres' definition of Regular which is a $T_{1}$ and $T_{3}$ space.
For a proper proof you need to go like this:-
Let $y_{1},y_{2}\in Y$ . Then $K_{1}=p^{-1}(y_{1})$ and $K_{2}=p^{-1}(y_{2})$ are compact sets and since we are in a Hausdorff space they are also closed.
Let $x\in K_{1}$ and $x'\in K_{2}$ . There exists $U_{x,x'}$ and $V_{x,x'}$ such that they are disjoint open sets containing $x$ and $x'$ respectively.
Then consider the open cover of $K_{1}$ by $\{U_{x,x'}\}_{x\in K_{1}}$ . This has a finite subcover $\{U_{x_{1},x'},...,U_{x_{n},x'}\}$ . Correspondingly we have the collection $\{V_{x_{1},x'},...,V_{x_{n},x'}\}$ . Let $\displaystyle U_{x'}=\bigcup_{j=1}^{n}U_{x_{j},x'}$ and $\displaystyle V_{x'}=\bigcap_{j=1}^{n}V_{x_{j},x'}$.
Then for each $x'\in K_{2}$ we have open sets $U_{x'}$ and $V_{x'}$ such that $x\in V_{x'}$ , $K_{1}\subset U_{x'}$ and $U_{x'}\cap V_{x'}=\phi$ .
Thus we consider the cover of $K_{2}$ by $\{V_{x'}\}_{x'\in K_{2}}$ . This has a finite subcover $V_{1},...,V_{m}$ and correspondily open sets $U_{1},...,U_{m}$. Then if we define $\displaystyle V=\bigcup_{i=1}^{m}V_{i}$ and $\displaystyle U=\bigcap_{i=1}^{m}U_{i}$ . Then $U$ and $V$ are open sets such that $K_{1}\subset U$ and $K_{2}\subset V$ and $U\cap V=\phi $.
Now we take the open sets $ W=Y\setminus p(X\setminus U)$ and $W'=Y\setminus p(X\setminus V)$. Then $y_{1}\in W$ and $y_{2}\in W'$ and $W\cap W'=\phi$ .
This proves Hausdorffness of $Y$.
Best Answer
Yes, I think. We only need closedness and continuity of $f$: $Y$ is normal (in the usual closed separation sense (or closed shrinkings), without $T_1$-ness assumed) by closedness of $f$; this is standard (let $\{U,V\}$ be an open cover of $Y$, then $\{f^{-1}[U], f^{-1}[V]\}$ is an open cover for $X$, so by normality has a closed shrinking $\{F,G\}$ with $F \subseteq f^{-1}[U], G \subseteq f^{-1}[V]$ and then closedness of the map gives us that $\{f[F], f[G]\}$ is a closed shrinking of $\{U,V\}$ as required).
Now if $U$ is open in $Y$, $f^{-1}[U]$ is a an $F_\sigma$ in $X$ (being open in a perfectly normal space) and so its image $U$ (by closedness and surjectivity of $f$) is also an $F_\sigma$. Now the usual Urysohn lemma proof applies to show $Y$ is perfectly normal in your sense as well.