Showing the image of a surjective map from a regular topological space is also regular Problem : Suppose that $X$ is a regular topological space, $Y$ a topological space. If there exists a continuous surjective map $f$ from $X \rightarrow Y$ that is closed, where $f^{-1}(y)$ is compact for every $y \in Y$, then I want to show that $Y$ is regular as well.
Thoughts:
First, I was thinking to use the characterization of a regular space that :
a topological space $X$ is regular iff for every $x \in X$, and every neighbourhood $U$ of $x$, there exists a neighbourhood $V$ such that $x \in V \subset cl(V) \subset X$.
So for $Y$ in this case take a point $y \in Y$, then $f^{-1}(y)$ is non-empty by surjectivity. Take $x$ in $f^{-1}(y)$, and then by the regularity of $X$ there are neighbourhoods $U,V$ of $x$ such that $x \in V \in cl(V) \in U$ where $f(cl(V))$ is a closed neighbourhood of $y$, say $C$. This is about as far as I get and I am not sure how to proceed.
 A: I will make use of the following two facts about closed continuous mappings.
Fact. Suppose that $f : X \to Y$ is a closed continuous map.


*

*Then given $y \in Y$ and an open $U \subseteq X$ with $f^{-1} ( y ) \subseteq U$ there is an open $W \subseteq Y$ with $y \in W$ and $f^{-1} ( W ) \subseteq U$.

*Then $\overline{ f(A) } = f ( \overline{A} )$ for all $A \subseteq X$.


(In fact, both of the above are equivalent characterizations of closed mappings.)

Pick $y \in Y$ and an open neighbourhood $V \subseteq Y$ of $y$.  Then $K = f^{-1} ( y ) \subseteq X$ is compact and $U = f^{-1} ( V ) \subseteq X$ is open and $K \subseteq U$.
By regularity of $X$ for each $x \in K$ there is an open $U_x \subseteq X$ with $x \in U_x \subseteq \overline{U_x} \subseteq U$. As $\{ U_x : x \in K \}$ is clearly an open cover of $K$, by compactness of $K$ there are $x_1 , \ldots , x_n \in K$ such that $K \subseteq U_{x_1} \cup \cdots \cup U_{x_n}$. Note that $\overline{U_{x_1} \cup \cdots \cup U_{x_n}} = \overline{U_{x_1}} \cup \cdots \cup \overline{U_{x_n}} \subseteq U$.
Since $f$ is closed using Fact(1) above there is an open $W \subseteq Y$ with $y \in W$ and $f^{-1} ( W ) \subseteq U_{x_1} \cup \cdots \cup U_{x_n}$. 
Using Fact(2) above (and the surjectivity of $f$) it follows that $$\overline{W} = \overline{f(f^{-1} (W)} = f ( \overline{f^{-1}(W) } ) \subseteq f ( \overline{ U_{x_1} \cup \cdots \cup U_{x_n} } ) \subseteq f ( U ) = V.$$
That is, we have found an open $W \subseteq Y$ such that $y \in W \subseteq \overline{W} \subseteq V$.
