Let $X$ be a topological space. For any $A\subset X$ we have $$\overline A = \bigcap\{ C:\ C\text{ is closed and }A\subset C\}.$$
Suppose that $X$ is Hausdorff, is it true that $$\overline A = \bigcap\{\overline U:\ U\text{ is a nbh of }A\}?$$ Here a neighborhood of $A$ simply means an open set containing $A$.
The above relation holds provided that $X$ is regular. I believe that it is false if we only assume $X$ to be Hausdorff but couldn't think of a counterexample.
As a prelimenary let me recall that regular means "a closed subset and a point outside it can be separated by open neighbourhoods". I will not assume that regular is Hausdorff. And indeed, the antidiscrete space shows that a regular space need not be Hausdorff. Also note that here $T_3$ means regular + Hausdorff.
Next if we look at the property then the "$\subseteq$" inclusion always holds regardless of separation axioms so actually we are only interested in the "$\supseteq$" inclusion.
Now you don't have to look far for a counterexample since:
Lemma. Being regular (not necessarily Hausdorff) is equivalent to the property.
Proof.
"$\Rightarrow$"
Assume that $A\subseteq X$ and $x\in \bigcap\{\overline{U}\ |\ U\text{ nbh }A\}$. Assume that $x\not\in\overline{A}$. Since $X$ is regular then there is a neighbourhood $U$ of $\overline{A}$ and a neighbourhood $V$ of $x$ such that $U\cap V=\emptyset$. Since $X-V$ is closed and $U$ is contained in it then $\overline{U}\subseteq X-V$ and this means that $x\not\in\overline{U}$. Note that $U$ is also an open neighbourhood of $A$ and therefore $x\not\in \bigcap\{\overline{U}\ |\ U\text{ nbh }A\}$. Contradiction. $\Box$
"$\Leftarrow$"
Assume that $X$ is not regular so we have a closed subset $A$ and a point $x\not\in A$ such that they cannot be separated. So any open neighbourhood $U$ of $A$ and any open neighbourhood $V$ of $x$ intersect. If we fix $U$ then by the arbitrary choice of $V$ we get that that $x\in\overline{U}$. And by the arbitrary choice of $U$ we get $x\in\bigcap\{\overline{U}\ |\ U\text{ nbh }A\}$. But $x\not\in A$ meaning
$$\bigcap\{\overline{U}\ |\ U\text{ nbh }A\}\not\subseteq A=\overline{A}$$
Contradiction. $\Box$
So for a counterexample all you have to do is to pick your favourite $T_2$ space that is not $T_3$.
Consider the real line $\mathbb{R}$ but with a non-standard topology generated by all subsets of the form $(a,b)-C$ where $C$ is a countable subset.
First of all this space is Hausdorff. If $a, b$ are two distinct reals then take $\epsilon:=|b-a|/2$ in order to separate them via $(a-\epsilon,a+\epsilon)$ and $(b-\epsilon, b+\epsilon)$. Note that intervals are open since the empty set is countable.
Now consider the irrationals $\mathbb{IQ}=\mathbb{R}-\mathbb{Q}$. Since rationals are countable and $\mathbb{R}$ is a union of open intervals then $\mathbb{IQ}$ is open and so $\mathbb{Q}$ is closed. Pick $r\in\mathbb{IQ}$. Then $r$ and $\mathbb{Q}$ cannot be separated. Indeed take any open neighbourhood $U$ of $r$. Then there is a countable subset $C$ such that $r\in(a,b)-C\subseteq U$. Now take some rational $q\in (a,b)$. Similarly any open neighbourhood of $q$ contains some subset of the form $(a_q,b_q)-C_q$. We can choose $a_q,b_q$ to be close enough to $q$ so that $(a_q,b_q)\subseteq (a,b)$. And in this situation
$$\big((a,b)-C\big)\cap\big((a_q,b_q)-C_q\big)=(a_q,b_q)-(C\cup C_q)$$
is nonempety by the cardinality argument. This shows that $r$ and $\mathbb{Q}$ cannot be separated.
For a more visual and less technical example please see this site:
http://mathonline.wikidot.com/a-t2-space-that-is-not-a-t3-space
