Ok, let's try another counterexample.
For simplicity, say a space with your property (any closed set is the intersection of its neighborhoods) is "Cromarty".
Let $X$ be an infinite set with the cofinite topology (the closed sets are all the finite sets and $X$ itself). If $Z \subset X$ is closed (hence finite), let $\mathcal{U}$ be the collection of all open sets containing $Z$. For any $x \notin Z$, $\{x\}^c$ is open (because it is cofinite) and contains $Z$, so $\{x\}^c \in \mathcal{U}$, and thus $x \notin \bigcap \mathcal{U}$; hence $Z^C \subset \bigcap \mathcal{U}$, so $\bigcap \mathcal{U}=Z$. So $X$ is Cromarty.
However, $X$ is not regular, since every pair of nonempty open sets has nonempty (in fact, infinite) intersection.
Edit: To amplify, it looks like the Cromarty property is equivalent to being $R_0$, which means: if $x$ has a neighborhood not containing $y$, then $y$ has a neighborhood not containing $x$. This is of course strictly weaker than being regular (as the cofinite topology shows).
Suppose $X$ is $R_0$, and $Z$ is closed. Suppose $x \notin Z$, $y \in Z$. Then $Z^c$ is a neighborhood of $x$ not containing $y$, so $y$ has a neighborhood $U_y$ that does not contain $x$. Now $U = \bigcup_{y \in Z} U_y$ is a neighborhood of $Z$ that does not contain $x$. As above, it follows that $Z$ is the intersection of its neighborhoods, so $X$ is Cromarty.
Conversely, suppose $X$ is Cromarty. Suppose a point $x$ has a neighborhood $U$ not containing $y$. $U^c$ is closed, hence the intersection of its neighborhoods, so $U^c$ has a neighborhood $V$ not containing $x$. But $V$ is also a neighborhood of $y$, so $X$ is $R_0$.
If $X$ is completely regular (closed set and point can be separated by a continuous real-valued function), then it is also regular. Let $x\in X$ be a point and $A\subset X$ be a closed set not containing $x$, then there is a continuous function $f:X\to\Bbb R$ with $f(x)=0$, $f(A)=1$, so $f^{-1}([0,\frac12))$ and $f^{-1}((\frac12,1])$ are disjoint open neighborhoods of $x$ and $A$, respectively.
To show that a regular $T_0$-space $X$ is $T_2$, let $x,y\in X$. There is a closed set $A$ containing one point, say $y$, but not the other. By regularity, there are disjoint open sets $U\ni x$ and $V\supseteq A\ni y$. So $X$ is Hausdorff.
Best Answer
Claim. A topological space $X$ has no regular opens other than $X$ and $\emptyset$ if and only if there is no pair of non-empty disjoint opens in $X$.
Proof. Left to right, if there are no regular opens other than $X$ and $\emptyset$, then for each non-empty open $U$ we have $\mathrm{cl}(U) = X$, since otherwise $\emptyset \subsetneq U \subseteq \mathrm{int} (\mathrm{cl} (U)) \subsetneq X$. In other words, $\mathrm{int}(X \setminus U) = \emptyset$, so there is no non-empty open $V$ disjoint from $U$. Right to left, if there is no pair of disjoint opens, then for each non-empty open $U$ we have $\mathrm{int}(X \setminus U) = \emptyset$, so $\mathrm{cl}(U) = X$ and $\mathrm{int}(\mathrm{cl}(U)) = X$. Thus $U$ is not regular unless $U = X$.
Non-example. No regular space which is not indiscrete satisfies this condition: if $F$ is a non-empty and non-total closed set, then regularity produces disjoint non-empty opens.
Non-example. No Hausdorff space with at least two points satisfies this condition: if $x$ and $y$ are distinct points, then the Hausdorff axiom produces disjoint non-empty opens.
Example. Each space where the specialization preorder is upward directed satisfies this condition. For example, the Alexandrov topology of an upward directed partially ordered set satisfies this condition, as does the topology on the reals consisting of intervals of the form $(a, +\infty)$ plus the empty and total set.