Can I do the following proof to show that the closure of an open ball is contained in the closed ball.
Let me pick y\in \bar(B(x,r)) in the closure. Then I know that this is equivalent to say that for all $\epsilon >0$ B(y,\epsilon)\cap B(x,r)\neq \emptyset. but this means that y\in B(x,r)\subset \bar(B)(x,r).