Define $A=\{(x,y)\,|\,x^2 \leq y \leq x\} \subset \mathbb{R}^2.$
How to prove that this subset is a closed set in $\mathbb{R^2}?$ Also, how to find the interior points?
To prove that $A$ is a subset I tried:
Let $((x_n,y_n))_{n \in \mathbb{N}}$ be a sequence with $(x_n,y_n) \to (x,y)$ in $\mathbb{R^2}$ for $n \to \infty.$
Then it's $x_n \to x, \ y_n \to y$ in $\mathbb{R}$ for $n \to \infty.$
Since $(x_n)^2 \leq y_n \leq x_n$ it's $x^2 = \left(\lim\limits_{n\to\infty}x_n\right)^{\!2} \leq \lim\limits_{n\to\infty}y_n = y \leq \lim\limits_{n\to\infty}x_n=x$
$\implies (x^2,x,y) \in A$ and this is why $A$ is a closed set. Does it work?
For the interior points I don't know a method to find them. But since $A^{\mathrm{o}}:=A$ \ $\partial A,$ I got
$A^{\mathrm{o}} = \{(x,y)\,|\,x^2 \leq y <x\}.$
Your argument for showing it's closed is correct.
For the interior, we know that any interior point has some ball around it that is fully contained in $A$. This can happen only when that point is not a limit of a sequence in the complement of $A$, so $(x_n)^2 > y_n$ or $y_n > x_n$ will have a limit that satisfies $x^2 \geq y$ or $y\geq x$, which means that $$A^\circ=\{(x,y)|x^2 < y <x\}.$$
For your guess you might note that $\partial A=\{(x,y)|x^2 = y \text{ or } y=x\}$, so you forgot a part of the boundary.