Easy topology inquiry 
(DEF 1): A topological space $X$ is a set with a collection $B$ of
  neighborhoods $N \subseteq X$ such that 
  
  
*
  
*for all $x \in X$, there exists $N \subseteq B$ such that $x \in N$
  
*for all $N_1, N_2 \in B$ with $x \in N_1 \cap N_2$, we have an $N_3 \in B$ such that $x \in N_3 \subseteq N_1 \cap N_2 $
$B$ is called a basis for the topology on $X$
(DEF 2): Let $X$ be a metric space. The metric topology on $X$ is
  defined by using set $N$ as neighborhoods of a point $x \in X$ where
$$D_X(x,r) = \{ y \in X : d(x,y) < r \} $$
where $r > 0$

(PROBLEM): Let $X$ be a metric space. Show that the collection of neighborhoods defined in (DEF 2) forms a basis for the metric topology.
SOLUTION: Take $x \in X$, and put $N = D_X(x,r)$. Then $x \in N$ since $d(x,x) = 0 < r$. So, condition (1) of def1 is satisfied. Now, put $N_1 = D_X(x_1,r_1)$ and $N_2 = D_X(x_2,r_2)$. Pick a point $x \in N_1 \cap N_2$. Therefore, we have $d(x_1, x) < r_1$ and $d(x_2,x) < r_2$. Now, put $N_3 = D_X(x_3, r_3)$ and let $r_3 = \min(r_1,r_2)$. Hence, any point in $N_3$ must satisfy $d(x_3,x) < \min (r_1,r_2) \leq r_1$. In other words, $x$ must lie in $N_3$ and $N_3 \subseteq N_1 \cap N_2 $. So, condition (2) of def1 is satisfied and the problem is solved.
Is this correct? I do not know If im doing what the problem is asking for. Can you give me some feedback? thanks a lot!

Here is the picture I was talking about. Take it as a hint:

