Question: My definition of $A\subset B$ dense in $B$ if $$\forall x\_0\in B, \forall \varepsilon >0, \mathcal B(x\_0 ,\varepsilon )\cap A\neq \emptyset.$$
In this definition, shouldn't we rather take $$\mathcal B(x\_0,\varepsilon )\setminus \{x\_0\}\neq\emptyset \ \ ?$$

Answer: 
No, the definition you gave is correct. Take for example $\mathbb N$ with discrete metric. Of course $\mathbb N$ is dense in it self. If you use $\mathcal B(x\_0,\varepsilon )\setminus \{x\_0\}$ instead of $\mathcal B(x\_0,\varepsilon )$, then when $\varepsilon <1$ and $n\in\mathbb N$, you'll get $$\mathcal B(n,\varepsilon )\setminus \{n\}=\emptyset,$$
what contradict density.
To rephrase the definition given,
$$
 \forall x\_0 \in B,\ \forall \varepsilon > 0,\ \exists a \in A,\ d(x\_0,a) < \varepsilon
$$
where $d$ is the metric. If $x\_0 \in A$ already, then that's fine. But what's more interesting is that points of $B \setminus A$ also have points of $A$ arbitrarily close to them.