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?


Just do a direct proof:
Let $\operatorname{diam}(A)$ be finite, and define $R = \operatorname{diam}(A)+1 > 0$. Pick $a \in A$. 
Then take any $x \in A$, then $d(x,a) \le \operatorname{diam}(A) < R$ (as $x,a \in A$ they are bounded by the $\sup$ of all distances between points of $A$, i.e. $\operatorname{diam}(A)$). This shows that $A \subseteq B(a,R)$ so $A$ is bounded.
