Proof of Open Ball in a Metric Space is Open

Let $y\in B_d(x,r)$. Then $d(x,y)<r$. Set

$$\varepsilon=r-d(x,y).$$

Since $d(x,y)<r$, one has $\varepsilon>0$. We claim that

$$B_d(y,\varepsilon)\subseteq B_d(x,r).$$

Indeed, if $z\in B_d(y,\varepsilon)$, then $d(y,z)<\varepsilon$. By the triangle inequality from Metric Space,

$$d(x,z)\le d(x,y)+d(y,z)<d(x,y)+\varepsilon=r.$$

Thus $z\in B_d(x,r)$.

So every point $y\in B_d(x,r)$ has an open ball centered at $y$ contained in $B_d(x,r)$. By Open Subset of a Metric Space, the set $B_d(x,r)$ is open in the metric space $(X,d)$.