The interior of a set $S$ is the set of points $s$ in $S$ for which you can find an open set $O$ such that $s \in O \subset S$.
So let $x_0$ be a point in $(0,1)$ and let $O$ be an open set containing $x_0$. Then by the definition of the Zariski topology you can write $O$ as $O = \bigcup_{p \in P} p^{-1} (\mathbb R \setminus \{0\})$ where $P$ is a subset of $\mathbb R [x]$, the set of polynomials with real coefficients. Since this means that $x_0 \in p^{-1} (\mathbb R \setminus \{0\})$ for some $p$ in $P$ so it's enough to find one polynomial $p$ such that $p(x_0) \neq 0$ and $p(x) = 0$ on all of $\mathbb R \setminus (0,1) = (-\infty , 0] \cup [1,\infty)$. The second condition needs to hold in order for $p^{-1}(\mathbb R \setminus \{0\})$ to be contained in $(0,1)$.
But if $p(x) = 0$ at any point $z$ then by continuity we can find a $\delta$ ball $B(z,\delta)$ on which $p(x) = 0$ which implies that $p$ is the zero polynomial. Hence we get a contradiction to $p(x_0) \neq 0$ and conclude that there can be no open sets contained in $(0,1)$ and hence its interior has to be empty.
For the closure: note that the closure of a set $S$ is all points $x_0$ such that for every open set $O$ containing $x_0$, $O \cap S \neq \varnothing$.
Now let $x_0$ be any point in $\mathbb R$ and let $O$ be an open set containing $x_0$. Then $O = \bigcup_{p \in P} p^{-1} (\mathbb R \setminus \{0\})$ for some set $P \subset \mathbb R [x]$. Assume $O$ does not intersect with $(0,1)$. Then all $p \in P$ are zero on $(0,1)$. But then again using continuity we get that all $p$ in $P$ have to be the zero polynomial. Hence $O = \mathbb R$ and hence $O \cap (0,1)$ is non-empty. Since we chose $x$ in $\mathbb R$ arbitrary we get that the closure of $(0,1)$ is all of $\mathbb R$.