If $r+r’\leq d(a,a’)$ then $B(a,r)\cap B(a’,r’)= \emptyset$. I have to prove this assertion : 
Suppose that $r+r’\leq d(a,a’)$ 
Show that $B(a,r)\cap B(a’,r’)= \emptyset$.
What I have to prove is that if I have $p \in B(a,r)\cap B(a’,r’)  $ then $p=0$. 
If $p \in B(a,r)\cap B(a’,r’)$ then : 
$$d(a,p)<r $$
$$d(a’,p)<r’$$
Then : 
$$0\leq d(a,p)+d(a’,p)\leq r+r’ \leq d(a,a’) \leq d(a,p)+d(p,a’)$$
Know I’m stuck, can I Say that $d(a,a’)=d(a,p)+d(a’,p)$ according to the last inequality ? If yes, I don’t know I to prove that $p=0$...


"What I have to proove is that if I have $p \in B(a,r)\cap B(a’,r’)  $ then $p=0$."

That is a wrong route and can only be marked as an effort to prove the weaker statement : $$B(a,r)\cap B(a’,r’)=\varnothing\text{ or } B(a,r)\cap B(a’,r’)=\{0\}$$
It must be shown that the assumption $p \in B(a,r)\cap B(a’,r’)$ leads to a contradiction.
If $p \in B(a,r)\cap B(a’,r’)$ then  $$d(a,a')\leq d(a,p)+d(p,a')<r+r'$$ contradicting that $r+r'\leq d(a,a')$.
