Hausdorffness of $\exp(X)$ implies regularity of $X$ I have trouble proving this fact about the exponential topology 

If $\exp(X)$ is Hausdorff and $X$ is T1, then $X$ is T3. 

What I need to show is that if I have a (i.e. closed) $\{x\} \subset X$ and a (disjointly taken to this point) closed set $A \subset X$, then I can separate them disjointly by open sets.
Does anybody now how to do this and where I get my open sets for $A$ and $\{x\}$ from?
 A: Just to make everything a bit more self-contained, I'll include the necessary definitions.
Notation. Let $X$ be a topological space.  By $2^X$ we denote the family of all nonempty closed subsets of $X$.1  Given $A_1 , \ldots , A_n \subseteq X$, we denote $$\langle A_1 , \ldots , A_n \rangle = \{ F \in 2^X : F \subseteq {\textstyle \bigcup_{i \leq n}} A_i, ( \forall i \leq n ) ( F \cap A_i \neq \varnothing ) \}.$$
Definition. Given a topological space, the exponential topology (or Vietoris topology) on $2^X$ is the topology generated by the base consisting of all sets of the form $\langle U_1 , \ldots , U_n \rangle$ where $U_1 , \ldots , U_n$ are open subsets of $X$. We denote by $\exp(X)$ the set $2^X$ under this topology.
1We only include the nonempty closed subsets in this definition because $\varnothing$ causes problems.  In particular, $\varnothing$ would not be an element of any of the basic open sets to be described, and so the exponential topology will fail to be T$_1$, let alone Hausdorff.

Let $x \in X$ and let $F \subseteq X$ be a (nonempty) closed set not containing $x$.  Then $F \cup \{ x \}$ is a closed subset of $X$ which is distinct from $F$, and so by Hausdorffness of $\exp(X)$ there are disjoint (basic) open sets containing them, respectively. That is, there are open sets $U_1 , \ldots , U_k, V_1 , \ldots , V_\ell$ in $X$ such that 


*

*$F \cup \{ x \} \in \langle U_1 , \ldots , U_k \rangle$;

*$F \in \langle V_1 , \ldots , V_\ell \rangle$;

*$\langle U_1 , \ldots , U_k \rangle \cap \langle V_1 , \ldots , V_\ell \rangle = \varnothing$.


It can be shown that $$U = {\textstyle \bigcap} \{ U_i : i \leq k, F \cap U_i = \varnothing \}$$ is an open neighbourhood of $x$ which is disjoint from $$V = V_1 \cup \cdots \cup V_\ell \supseteq F.$$
