Closed balls are closed

Proposition. In any metric space, every closed ball $B'(x_0;r)$ is a closed set .

Proof sketch. Show that the complement of $B'(x_0;r)$ is open: if $x\notin B'(x_0;r)$ then $d(x,x_0)>r$. Let $\delta:=d(x,x_0)-r>0$. If $d(y,x)<\delta$, then $d(y,x_0)\ge d(x,x_0)-d(y,x)>r$, so $y$ also lies outside the closed ball. Hence a ball around $x$ lies in the complement, proving openness of the complement.