Radius of an open ball

Question

Let $(X,d)$ be a metric space.

Show that if $q\in B(x,r)$ and $q\neq x$ then there exists $0<R<d(q,x)$ such that $B(q,R)\subseteq B(x,r)$.

I was wondering as to how to approach it:

Since $B(x,r)$ is open, there exists a radius $r'>0$ such that $B(q,r')\subseteq B(x,r)$. But this doesn't necessarily mean that $r'\leq r$. I know that there exists an $R$ such that $0<R<d(q,x)$. But otherwise, I'm not sure. I highly suspect that it has to do with $X$ being $T_1$.

Answer 1

Hint: Let $0<R<r-d(q,x)$ and conclude by applying the triangle inequality.