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 $p=0$...

I assume the balls are open, otherwise the statement is not true.
Let $p\in B(a,r)$, i.e. $d(a,p)\lt r$. Then the triangle inequality states that
$$d(a,p)+d(p,a')\ge d(a,a')\ge r+r'$$
Hence
$$d(p,a')\ge (r-d(a,p))+r'>r'$$
Hence $p\notin B(a',r')$.
Therefore, $B(a,r)\cap B(a',r')=\emptyset$.

Note that it's absolutely not true that you have to show that if $p \in B(a,r)\cap B(a’,r’)$ then $p=0$. If $p=0$ were in $B(a,r)\cap B(a’,r’)$, then obviously the intersection would not be empty.
