Let $$A:=\left\{(x,y)\in\mathbb R^2:y\le 2x\text{ and }y\le-x\right\}.$$
How can we determine the boundary $\partial A$ of $A$? Intuitively, from the plot, the boundary should consists of the two infinite line segments starting at $0$. But how can we show this formally?
Dropping the $(x,y) \in \mathbb{R}^2$ part, we may write $$A \supset \{y < 2x \text{ and } y < -x\} = \{y-2x<0\} \cap \{y+x<0\}$$ which is open since $y-2x$ and $y+x$ are continuous as functions of $x$ and $y.$ Similarly, we have $$\mathbb{R}^2 \setminus A = \{y-2x>0\} \cup \{y+x > 0\}$$ which is open.
We conclude that $$\partial A \subseteq A \setminus \left(\{y-2x < 0\} \cap \{y+x<0\}\right) = \left(\{y=2x\}\cap\{y\leq -x\}\right)\cup\left(\{y\leq 2x\}\cap\{y=-x\}\right)$$
To show the other direction, let $(x,y)$ be some point from the set represented on the right-hand side and note that every open ball around $(x,y)$ contains a point of $A$ and a point of $\mathbb{R}^2\setminus A.$
Here, I'll let you fill in the details for this last step, as it's straight-forward, if not a little tedious in rigor.