Let $(X,d)$ be a metric space and $A \subseteq X$. Let $B(x,r)$ be the open ball with center $x \in X$ and radius $r>0$. The set $A$ is closed, $diam(A)<r$ and also $A \cap B(x,r) \neq \emptyset$
Prove that $A \subseteq B(x,2r)$.
If $x \in A$ then it's rather obvious that any ball $B(x,r)$ contains the set $A$ so $A \subseteq B(x,2r)$. Now it remains to prove for $x \in X \setminus A$. From the condition $A \cap B(x,r) \neq \emptyset$ there is always a point from $A$ that is in the ball at the same time so at worst we will have that point be $y \in A$ such that $d(z,y) \leq r$ $\forall z \in A$ in which case the ball $B(x,2r)$ always contains the set $A$.
This is a very informal proof and it mostly comes from intuitive understanding. I'm not sure how to write a more formal one using the definitions of closed set, ball and the conditions. Any help is appreciated.
Let $y\in A\cap B(x,r)$. Then, for every $a\in A$, $$ d(a,x) \leq d(a,y) + d(y,x) < r+r=2r, $$ therefore $A\subset B(x,2r)$.