Similarly to the other answers, we will need to compare the usual distance of $\mathbb{C}^n$ and $\delta$ (which, as you noted, does not induce a distance). Let $m=\inf\left\{\delta(z):\Vert z\Vert=1\right\}$ and $M=\sup\left\{\delta(z):\Vert z\Vert=1\right\}$. The condition on $\delta$ implies
$$m\Vert z\Vert\leq\delta(z)\leq M\Vert z\Vert$$
for all $z\in \mathbb{C}^n$. Note that the first inequality implies that whenever $(z_n)$ is a sequence in $\mathbb{C}^n$ with $\Vert z_n\Vert\to\infty$, we also have $\delta(z_n)\to\infty$ (these are sometime called proper functions, I believe).
Lemma: For every $z\in\mathbb{Z}^n$, there exists $w\in\Omega^c$ such that $\delta(z,\Omega^c)=\delta(z-w)$. Moreover, $\Vert w\Vert\leq(1/m)\delta(z,\Omega^c)+\Vert z\Vert$.
Proof: Let $w_n$ be a sequence in $\Omega^c$ with $\delta(z-w_n)\to \delta(z,\Omega^c)$. Then $(w_n)$ is bounded by the comment above, so we can take a converging subsequence and obtain such $w$. Then
$$\Vert w\Vert\leq\Vert z-w\Vert+\Vert z\Vert\leq (1/m)\delta(z-w)+\Vert z\Vert=(1/m)\delta(z,\Omega^c)+\Vert z\Vert.QED$$
Now let $z\in \mathbb{C}^n$ be fixed. As you noted, $\delta(\cdot,\Omega^c)$ is lower semicontinuous, so there exists $r>0$ such that $\delta(z',\Omega^c)<\delta(z,\Omega^c)+1$ whenever $\Vert z'-z\Vert\leq r$. For all such $z'$, the lemma gives us $w'\in\Omega^c$ with $\delta(z',\Omega^c)=\delta(z'-w')$ and $\Vert w'\Vert\le(1/m)\delta(z',\Omega^c)+\Vert z'\Vert\leq(1/m)\delta(z,\Omega^c)+1+\Vert z\Vert+ r$.
Let $\overline{B_r(z)}$ be the closed ball of radius $r$ around $z$ and $K=\Omega^c\cap\left\{w:\Vert w\Vert\leq(1/m)\delta(z,\Omega^c)+1+\Vert z\Vert\right\}$. The previous paragraph implies that
$$\delta(z',\Omega^c)=\inf_{w\in K}\delta(z'-w),\qquad z'\in \overline{B_r(z)}$$
Now things are easy, since we can work in the compact $K$ instead of all $\Omega^c$. Precisely, we just need to prove the following proposition:
Proposition: Let $X$ and $Y$ be compact metric spaces and $f:X\times Y\to \mathbb{R}$ be continuous. Define $F(x)=\inf_{y\in Y}f(x,y)$ for all $x\in X$. Then $F$ is continuous.
Our problem follows from this by letting $X=\overline{B_r(z)}$, $Y=K$ and $f(x,y)=\delta(x-y)$, so that $\delta(\cdot,\Omega^c)=F$ in $\overline{B_r(z)}$.
Proof of proposition: Since $f$ is a continuous function on a compact, it is uniformly continuous. Given $\epsilon>0$, there exists $\nu>0$ such that $d(x,x')<\nu$ and $d(y,y')<\nu$ implies $|f(x,y)-f(x',y')|<\epsilon$. Suppose $x,x'\in X$ with $d(x,x')<\nu$, and we will show that $|F(x)-F(x')|<\epsilon$.
Choose $y$ and $y'$ in $Y$ with $f(x,y)=F(x)$ and $f(x',y')=F(x')$. Then the choice of $\nu$ implies $|f(x',y)-f(x,y)|<\epsilon$ and $|f(x,y')-f(x',y')<\epsilon$, thus
$$F(x')=f(x',y')>f(x,y')-\epsilon\geq F(x)-\epsilon$$
and
$$F(x)=f(x,y)>f(x',y)-\epsilon\geq F(x')-\epsilon$$
therefore $|F(x')-F(x)|<\epsilon$, as we wanted. QED
This shows that $\delta(\cdot,\Omega^c)$ is continuous on $\overline{B_r(z)}$, so in particular it is continuous at $z$.
