Open balls are open

Proposition. In any metric space , every open ball $B(x_0;r)$ is an open set .

Proof sketch. Fix $a\in B(x_0;r)$. Let $\delta:=r-d(a,x_0)>0$. If $d(x,a)<\delta$, then by the triangle inequality

so $x\in B(x_0;r)$. Hence $B(a;\delta)\subset B(x_0;r)$, which is exactly openness.