We say that a subset $A$ of $\mathbb{R}$ is approved when $\forall x,y \in \mathbb{R}$ if $y \in A$ and $x \leq y$ then $x \in A$. I want to find all the approved subsets of $\mathbb{R}$.
Let $x,y \in A$ with $x \leq y$ and $y \in A$. If $x \in A$, doesn't it mean that the set of the $x$s is a subset of the set of $y$s.
But does this help somehow?
On
Let $\alpha=\sup A$. Then $(-\infty,\alpha)\subset A\subset(-\infty,\alpha]$, so $A$ is one of those two intervals, depending on whether $\alpha\in A$.
(Note this includes the case $\alpha=+\infty$, in which case $A=(-\infty,\infty)=\Bbb R$.)
(And, as pointed out by @MPW, it also includes $\alpha=-\infty$, which leads to $A=\varnothing$.)
Your definition seems to say that if $$y\in A$$ then $$(-\infty,y]\subset A.$$ To me this says that $A$ is either empty, all of $\mathbb R$, a set of the form $(-\infty,a)$, or a set of the form $(-\infty,a]$, and that these are the only possibilities.
This is because $$A=\bigcup_{a\in A}(-\infty,a].$$