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.

This generates the same topology as the classical definition.
To see this, note that any closed ball $\overline{B_\epsilon(x)}$ of positive radius $\epsilon>0$ contains an open ball of positive radius, namely $B_{\epsilon\over 2}(x)$; and conversely as well ($B_\epsilon(x)$ contains $\overline{B_{\epsilon\over 2}(x)}$). So any set which is open in one sense, is open in the other.

Note that there's a related idea you might try: take the topology generated by closed balls of positive radius! Unfortunately, this does not give the usual topology, and in fact yields rather silly topology - the discrete topology, in which every set is open. 
Why? Well, any singleton $\{x\}$ is the intersection of two closed balls of positive radius (picture two spheres touching at one point), and since a topology must be closed under arbitrary unions, this means that every set is open: if $S$ is a set of points, then $S=\bigcup_{x\in S}\{x\}$ is a union of open sets, hence open.

Note that if in your definition you drop the requirement that the radius $r$ be positive, then every singleton is closed (since the closed ball of radius $0$ around $x$ is just $\{x\}$); so this again yields the discrete topology. So it's important to restrict attention to positive radii.
