Topological spaces in which a set is the support of a continuous function iff it is the closure of a open set.
Here is a counter-example (compact and Hausdorff).
I will use ordinal topology (see e.g. first uncountable ordinal and Willard's General topology). Here's a crash-course:
The first uncountable ordinal is an uncountable totally ordered set $\omega_1$ with the property that all of its downsets $\left\{x\in\omega_1:x<\alpha\right\}$ (where $\alpha\in\omega_1$) are countable. Such a set exists and is unique up to order isomorphism.
The second uncountable ordinal is the set $\Omega=\omega_1\sqcup\{\omega_1\}$, the union of $\omega_1$ with a new point. It is also an ordered set with $\omega_1=\max\Omega$. With the order topology, $\Omega$ is a compact Hausdorff, non-second countable space. Every continuous function from $\Omega$ to any first-countable space is constant on a neighbourhood of $\omega_1$.
Let $X_1$ and $X_2$ be two copies of the second countable ordinal with maxima $x_1$ and $x_2$, respectively, and let $X$ be the space obtained by gluing $x_1$ and $x_2$ (i.e., the quotient of the disjoint union $\Omega\times\left\{1,2\right\}$ by identifying $(\omega_1,1)\sim(\omega_1,2)$).
Call the image of $x_i$ in $X$ just $x_0$. Then $X$ contains $X_1$ as a regular closed subset. the interior of $X_1$ is just $\left\{x\in X_1:x<x_0\right\}$, and $X_1$ is the closure of this set. Similarly for $X_2$.
However, $X_1$ is not the support of any continuous function $f\colon X\to \mathbb{R}$. Indeed, if $X_1$ were the support of $f$, then $f=0$ on $X\setminus X_1$, which is the interior of $X_2$, so $f=0$ on $X_2$ as well. In particular $f(x_0)=0$. However $f$ is constant on a neighbourhood of $x_0$, by general properties of the first uncountable ordinal (and because $\mathbb{R}$ is first countable). So the support of $f$ is strictly smaller than $X_1$, a contradiction.
I guess that given a subset $Y$ of a (topological) space $X$ by $Y$ is a support of a continuous function you mean that there exists a continuous function $f:X\to \Bbb R $ such that $Y=\overline{f^{-1}(\Bbb R\setminus\{0\} )}$. In particular, $Y$ is the closure of an open set. More precisely, $Y$ is the closure of a functionally open set (see [E, 1.5.13] below). So if $X$ is a normal space then supports of continuous functions are exactly closures of open $F_\sigma$-subsets of $X$. Remark that by [E, 4.1.12] each open subset of a metric space is functionally open, which verifies your claim that for a metric space $X$, supports of continuous functions are exactly closures of open subsets of $X$. At last, in order to provide the required counterexample it suffices to find in a normal space $X$ and its subset $Y$ which is a closure of an open set (such sets are called regular closed sets), but not a closure of an $F_\sigma$-open set.
References
[E] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
As Ravsky wrote, in a normal space $X$, a set $E$ is the support of a continuous function if and only if $E$ is an open $F_\sigma$-set, and Luiz Cordeiro gave an example of a normal set $X$ and a compact set $X_1$ which is the closure of an open set, but is not $F_\sigma$, completing the answer. Moreover the claim
A set $E$ is the support of a continuous function if and only if $E$ is the closure of an open set
is true for perfectly normal spaces. If you allow less refined spaces, I found a pretty good example. Let $\mathbb{Q}$ be the rational numbers and $\mathbb{Q}^* $ its one point compactification (viz. $\mathbb{Q}^* = \mathbb{Q} \cup \{p\}$ for some element $p \notin \mathbb{Q}$ and the open sets $U$ in $\mathbb{Q}^* $ are the open sets of $\mathbb{Q}$ and the complement of compact sets of $\mathbb{Q}$, that is $U=\mathbb{Q}^* - K$ for some compact $K \subset \mathbb{Q}$). The important step in this proof is the following result:
Every compact set in $\mathbb{Q}$ has empty interior
whose proof can be read here. Since this space is compact (well, it is a compactification), then every closed set is compact. However, this space is not Hausdorff. Now, let $V$ be any open set in $\mathbb{Q}$ which is not dense, for example $V=(-\infty, 0) \cap \mathbb{Q}$. Then $\operatorname{cl}_{\mathbb{Q}^* }(V) = \operatorname{cl}_{\mathbb{Q}}(V) \cup \{p\}$. I claim this set is not the support of any function. In fact, let $f : \mathbb{Q}^* \to \mathbb{R}$ be a continuous function, and suppose $f(p)=\alpha$ and $f(r)=\beta$ for some $r \in \mathbb{Q}$ and $\alpha \neq \beta$. Then $\alpha$ and $\beta$ have disjoint neighborhoods $U$ and $V$, but then $f^{-1}(U)$ is a neighborhood of $p$ and it is the complement of a set with empty interior, and $f^{-1}(V)$ is an open set in $\mathbb{Q}$ and therefore there exists in interval such that $I \cap \mathbb{Q} \subset f^{-1}(V) \subset \mathbb{Q}^* - f^{-1}(U)$, which is impossible. Therefore $f$ must be constant! And the only supports of continuous functions are $\emptyset$ and the whole space $\mathbb{Q}^* $. Hence we have found an example of a compact $T_1$ space $\mathbb{Q}^* $ and a compact set which is the closure of an open $F_\sigma$-set (since $\mathbb{Q}$ is denumerable), but is not the support of a continuous function, showing that the claim above only follows in normal spaces.
(Obs.: We have also proved this space in not normal. In fact, it is a known fact that a space is normal if and only if Tietze Extension Theorem is true. On the other hand, if $A = \{0,1\}$ and $f : A \to \mathbb{R}$ is defined by $f(x)=x$ than $f$ is clearly continuous, but it does not have a continuous extension on $\mathbb{Q}^* $ since it is not constant.)