Is a straight line closed in $\mathbb{R}^2$ with no interior points?

Yes, you are right. Consider the line $L=\{(x,0):x\in\Bbb R\}$, i.e. the $x$-axis. If let $P=(x,0)$, then the ball $B(P,\epsilon)$ for any $\epsilon>0$ will contain points with $y>0$ and $y<0$. So, no $\epsilon$ ball is contained in the line.

Let denote $L$ a line in $\mathbb R^2$. We have that $$\forall x\in L,\forall \varepsilon>0,\quad B_x(\varepsilon)\cap L\neq\emptyset\quad\text{and}\quad B_x(\varepsilon)\cap L^c\neq\emptyset$$

where $B_x(\varepsilon)=\{y\in\mathbb R^2\mid \|x-y\|<\varepsilon\}$. Therefore $$L=\partial L,$$ and thus $L$ is closed with no interior point.

In fact, you can prove that if $(V,\|\cdot \|)$ is a vector space such that $\dim V=n$, then all subspace $W$ such that $\dim W\leq n-1$ is closed with no interior point in $V$.

Another way to see that it is closed is to check that the complement is open: Around any point not on the line, you can find an open ball not intersecting the line (e.g. the ball with radius half the distance of the point from the line).

And of course, an open ball around a point of the line will always contain points both from the line and from its complement, therefore every point of the line is a boundary point and its interior is indeed empty.