Let $A = \{(x,y): 5x+4y \leq 20, x \geq 2, y \geq 1\}$. With the usual Euclidean metric $d$, prove that $A$ is bounded and closed on $(X,d)$. Is $A$ compact?
My attempt: $A = \{(x,y): 2 \leq x \leq \frac{16}{5} , 1 \leq y \leq \frac{5}{2} \}$. So let $r_0 = \sqrt{(\frac{16}{5})^2 + (\frac{5}{2})^2 }$, choose $r> r_0$, then the open ball S with center $(0,0)$ and radius $r$ always contains $A$. Thus $A$ is bounded.
$\mathbb{R}^2 \backslash A = \{(x,y): x <2 \ \text{or} \ x> \frac{16}{5} , y <1 \ \text{or} \ y >\frac{5}{2} \}$. Since this set is open, $A$ is closed.
$A$ is compact because $\mathbb{R}^2$ is a complete space, then any bounded and closed set is compact.
Is the proof correct and rigorous enough?