Let $(X,d)$ a metric space and $A \subset X$. $A$ is bounded if and only if $\operatorname{diam}(A)$ is finite. Good morning, I have a problem to prove the following theorem.
Theorem:Let $(X,d)$ a metric space and $A \subset X$.  $A$ is bounded if and only if $\operatorname{diam}(A)$ is finite. 
Prove:

$->$ A is bounded, then exist $a \in X$ and $r>0$ such that $A\subset B(a,r)$
  Then $d(a,x)<r \forall x\in A$
  Then $d(x,y) : x,y \in A < d(x,a) + d(y,a) < r + r = 2r$
  In consequence we have:
  $d(x,y)\leq sup${$d(x,y):x,y \in A$}$<2r<$$\infty$.
   In other words: $\operatorname{diam}$ is finite.
  

I have problem with the other implication. 
If $\operatorname{diam}(A)$ is finite, then $A$ is bounded.
I think this:


Suppose $A$ is not bounded.
  
  Then $\forall a \in X$ and $r>0$ don't happen the following: $A \subset B(a,r)$
  Then $d(x,a)>r \forall x \in A$
  In this step i'm stuck. Can someone help me?



"Then $\forall a \in X$ and $r>0$ don't happen the following: $A\subset B(a,r)$.
  Then $d(x,a)>r \forall x \in A$."

This is not true, the negation of "$A\subset B(a,r)$" is "there exists $x\in A$ such that $x\notin B(a,r)$".
So, fixed $y\in A$, we have that for every $r>0$ there exists $x\in A$ such that $d(x,y)\geq r$. Therefore, it can't be true that
$${\rm diam}(A) = \sup\{d(x,y) : x,y\in A\} < \infty.$$
