Your solution is practically there. You're right that the evenly spaced nested balls is what's missing.
Fix $y \in \Omega$ and fix some multi-index $\alpha$ with $| \alpha | = k$. Let $d = dist(y, \partial \Omega)$. Let $d_{0} = \frac{d}{|\alpha|}$.
Then consider the ball $B = B(y, d_{0})$. Let $\alpha_{1}$ be a multi-index such that $|\alpha_{1}| = k-1$ and $\alpha_{1} < \alpha$. Based off the estimate you already know, we have that
$$
|D^{\alpha} u(y)| \le \frac{n}{d_{0}} \sup_{B(y,d_{0})} |D^{\alpha_{1}} u|.
$$
However, for every $y_{1} \in B(y, d_{0})$ we can apply the same estimate on the ball $B(y_{1}, d_{0})$ to gain
$$
|D^{\alpha} u(y)| \le \frac{n}{d_{0}} \sup_{y_{1} \in B(y,d_{0})} \left[ \frac{n}{d_{0}} \sup_{B(y_{1}, d_{0})} |D^{\alpha_{2}}u| \right] \le \frac{n}{d_{0}} \left[ \frac{n}{d_{0}} \sup_{B(y,2 d_{0})} |D^{\alpha_{2}} u| \right],
$$
where $|\alpha_{2}| = k-2$, and $\alpha_{2} < \alpha_{1}$.
Repeating inductively, we attain the desired result
$$
| D^{\alpha} u(y) | \le \left( \frac{n}{d_{0}} \right)^{|\alpha|} \sup_{B(y, |\alpha| d_{0})} |u| \le \left( \frac{n |\alpha|}{d} \right)^{|\alpha|} \sup_{\Omega} |u|,
$$
where the equality follows since we chose $d_{0} = \frac{d}{|\alpha|}$.
Best Answer
Because if $U$ is bounded and $f:U\to \mathbb R$ is uniformly continuous, then $\lim_{x\to y}f(x)$ exist for all $y\in \partial U$. The way to prove is the following :
If $(a_n)\subset U$ is s.t. $a_n\to y\in \partial U$, then $(f(a_n))_n$ is a Cauchy sequence. From this, you can get $\lim_{x\to y}f(x)$ exist.