Why does the definition of open set in Euclidean topology involve the open ball and not the closed ball? The definition of Euclidean topological space $( \Bbb{R}^n,\tau)$ in most standard texts is
$$\text{$\Omega$ is open iff $\forall$ $p$ $\in \Omega$ $\exists$ $r>0$ s.t. $D_r(p) \subset \Omega$ where $D_r(q):= \{x : |x-q|<r \}$ }$$
What exactly would  not work if the open ball $D_r(q)$ were defined with a $\leq$ sign? As in
$$\text{$\Omega$ is open iff $\forall$ $p$ $\in \Omega$ $\exists$ $r>0$ s.t. $D_r(p) \subset \Omega$ where $D_r(q):= \{x : |x-q|\leq r \}$ }$$
I have thought about this and haven't been able to reach any definite conclusion.

If we write $B_{r}(p)=\{x \in \mathbb{R}^{n}|\  |x-p|<r\}$, and $\bar{B_{r}}(p)=\{x \in \mathbb{R}^{n}|\  |x-p|\le r\}$, then for any $p \in \mathbb{R}^{n}$ and $r \in \mathbb{R}$, we have the inclusions $$\bar{B}_{r/2}(p) \subset B_{r}(p) \subset \bar{B_{r}}(p)$$
So let $S$ be a set and $p \in S$. If there is an open $r$-ball inside $S$ containing $p$ then there is a closed $r/2$-ball inside $S$ containing $p$, and similarly if there is a closed $r$-ball in $S$ containing $p$ then the open $r$-ball also contains $p$ and is inside $S$. In other words, a set is open in the usual sense if and only if it is open in your slightly different sense.  
However, note that closed balls are still not open! In particular, points on the boundary of the ball do not have the property you give (you might not have thought this, but it's worth warning in case you did!)
