Elementary question about topology and metric spaces Let $(X, \rho ) $ be a metric space. Denote
$$ \mathscr{B} = \{ B(\epsilon,x) : x \in X, \epsilon>0 \} $$
Let $B_1,B_2 \in \mathscr{B}$ . Let $x \in X$ be arbitrary with $ x \in B_1 \cap B_2 $. I want to find $B_3 \in \mathscr{B}$ such that $x \in B_3 \subset B_1 \cap B_2 $
Attempt:
Fix $b_1,b_2 \in X$ and we say they are the centers of $B_1,B_2$, respectively. Then we have  $B_1 = \{ z \in X : \rho(z, b_1 ) < \epsilon_1 \} $ and $B_2 = \{ z \in X : \rho(z, b_2 ) < \epsilon_2 \} $. Let $x \in X$ be arbitrary with $x \in B_1 \cap B_2 $. After doing some drawings, I think the correct choice for $\epsilon $ would be
$$ \epsilon = \frac{ \min\{ \epsilon_1 - \rho(x,b_1), \epsilon_2 - \rho(x,b_2 )\}}{2} $$
Now, put $B_3 = \{ y : \rho(x,y) < \epsilon \} $. We must check $B_3 \subset B_1 \cap B_2 $. To do this, let $y \in B_3 $. Next,
$$ \rho(y,b_1) \leq \rho(y,x) + \rho(x, b_1) < \epsilon + \rho(x,b_1) \leq \frac{ \epsilon_1 - \rho(b_1,x)}{2} + \rho(b_1,x) =  \frac{ \epsilon_1}{2} + \frac{ \rho(b_1,x)}{2} < \epsilon_1/2 + \epsilon_1/2 = \epsilon_1 $$
Therefore, $y \in B_1$. Similarly, we obtain that $y \in B_2$. In particular, $y \in B_1 \cap B_2$. This implies $B_3 \subset B_1 \cap B_2$.
I would like to know if this a correct approach. Maybe is too long or maybe it has flaws, I don't know. Any feedback would be greatly appreciated
 A: $x\in B_{i}=B\left(\epsilon_{i},x_{i}\right)=\left\{ z\in X\mid\rho\left(x_{i},z\right)<\epsilon_{i}\right\} $
for $i=1,2$. 
Let's choose for a $B_3$ that is centered at $x$, i.e.
$B_{3}=B\left(\epsilon,x\right)$ for some
$\epsilon>0$. 
To achieve $B_3\subset B_1\cap B_2$ it must be so that $\rho\left(x,z\right)<\epsilon$
implies $\rho\left(x_{i},z\right)<\epsilon_{i}$ for $i=1,2$. 
$\rho\left(x_{i},x\right)<\epsilon_{i}$ for $i=1,2$ makes it
possible to find some $\epsilon>0$ with $\epsilon<\epsilon_{i}-\rho\left(x_{i},x\right)$
for $i=1,2$. 
Then $\rho\left(x_{i},z\right)\leq\rho\left(x_{i},x\right)+\rho\left(x,z\right)<\rho\left(x_{i},x\right)+\left(\epsilon_{i}-\rho\left(x_{i},x\right)\right)=\epsilon_{i}$
for $i=1,2$.
A: You did everything right.
By the way, it is enough to define
$$ \epsilon:=\min\{ \epsilon_1 - \rho(x,b_1), \epsilon_2 - \rho(x,b_2)\}$$

This fact will be trivial when you study general topological spaces. In that case you could just say
$$B_1, B_2\in \tau \ \Rightarrow \  B_1\cap B_2 \in \tau\ \Rightarrow \forall x\in B_1\cap B_2\ \ \exists\, \epsilon>0 \text{ such that } x\in B(x,\epsilon)\subset B_1\cap B_2$$
A: I think your idea is basically right, but it would be better to label $b_1$ and $b_2$ the centres of the balls rather than calling them the "radiuses" which is just confusing. I think your proof would go easier that way. Also consider that a ball is just an open set, so any element within a ball $B$ must have another, smaller, ball around it which is completely contained in $B$. Then just take the smaller of the two radiuses of the balls. That is a much quicker route to the result.
