Is every open set the interior of a closed set?

The set $(0,1) \cup (1,2)$ is a counterexample.

No. A quick way to verify counterexamples is the following observation: if $U$ is the interior of a closed set $C$, then $U$ is also the interior of $\overline{U}$. Indeed, since $C$ is closed and $U\subseteq C$, $\overline{U}\subseteq C$, so the interior of $\overline{U}$ is contained in the interior of $C$. But $U$ is contained in the interior of $\overline{U}$ (since it is open and contained in $\overline{U}$) and equal to the interior of $C$, so $U$ must be equal to the interior of $\overline{U}$.

We can now give an example: take $U=(0,1)\cup(1,2)$ as a subset of $\mathbb{R}$. Since $\overline{U}=[0,2]$ has interior $(0,2)\neq U$, $U$ cannot be the interior of any closed set. In general, a set which is the interior of a closed set (equivalently, of its closure) is called a regular open set.

Since the complement of an open set is closed (and vice versa), and since the complement of the interior is the closure of the complement, we can rephrase your question equivalently as:

Is every closed set the closure of some open set?

This immediately suggests a counterexample: any singleton (i.e. a set containing only one point) is closed in $\mathbb R^n$ (with the usual Euclidean topology), but has no non-empty open subsets that it could be the closure of.

Conversely, the complement of any singleton (i.e. $\mathbb R^n \setminus \{x\}$ for any $x \in \mathbb R^n$) provides a counterexample to your original claim, being an open set that cannot be the interior of any closed set.