This question seems natural enough that the answer should be known, but I was unable to find a reference.
Call a subset $C$ of $\Bbb R$ left- (respectively, right-) perfect if it is perfect and if every point in $C$ is a limit point from the left (respectively, from the right). Are there any subsets of $\Bbb R$ that are Lebesgue-null, right-perfect and left-perfect?
Clearly, the standard Cantor set is neither right- nor left-perfect.
The only left-perfect sets are $\emptyset$, $\mathbb{R}$, and closed half-lines of the form $(-\infty, b]$. So the only sets which are right-perfect and left-perfect are $\emptyset$ and $\mathbb{R}$, and the only set fitting the criteria of your question is $\emptyset$.
To see this, suppose $y \notin C$. If there exists $x \in C$ with $x > y$, we can let $x_0 = \inf\{x \in C : x > y\}$. Then $x_0 \in C$ since perfect sets are closed, but there are no points of $C$ between $y$ and $x_0$, so $x_0$ is not a limit point from the left, a contradiction. Thus $[y, \infty) \subset C^c$. It follows that $C^c$ is a half-line which is open since $C$ is closed, so $C^c = (b, \infty)$ for some $b$, or else $C^c = \mathbb{R}$ or $\emptyset$.