real analysis - $log$ is continuous

I'm trying to prove that the complex logarithm function is continuous using this theorem, but I'm hitting a snag in part of the proof.

Let $D=\Bbb C\setminus(-\infty,0]$. The claim is that the function $\log\upharpoonright D:D\to\Bbb C$ is continuous on $D$. What we know so far is that $\log$ is defined as the inverse of $\exp\upharpoonright (\Bbb R\times(-\pi,\pi])$ (where the notation refers to the set of all complex numbers with real part in $\Bbb R$ and imaginary part in $(-\pi,\pi]$), and it is well-defined because we have shown that $\exp$ on this domain is a bijection onto $\Bbb C\setminus\{0\}$. We also know that $\exp$ is continuous.

Given $x\in D$, we wish to show that $\log$ is continuous at $x$. In order to apply the linked theorem, we need a compact region, so let $y=\log x$ and define $Y=[\Re y-1,\Re y+1]\times[\frac{\Im y-\pi}2,\pi]$. Then $Y$ is compact, so $\exp(Y)$ is also compact, and since $y\in \Bbb R\times(-\pi,\pi]$ follows from the definition of $\log$ and $\Im y\ne\pi$ because this would imply $x\in(-\infty,0]$, we also have $$y\in Y^\circ=(\Re y-1,\Re y+1)\times(\frac{\Im y-\pi}2,\pi).$$

Now we can apply the theorem to deduce that $\exp\upharpoonright Y:Y\to\exp(Y)$ is a homeomorphism, so $\log$ is continuous on $\exp(Y)^\circ$. Where I got stuck is in the last part, to show that $x\in\exp(Y)^\circ$ given that we already know $x\in\exp(Y^\circ)$, because the subspace topologies involved don't play well with interior here. Specifically, we know that $\exp(Y^\circ)$ is open in $\exp(Y)$, but I don't see how this implies that it is open in $\Bbb C$ (or $D$).

I realize that I can probably grok this proof with sufficient details of the shape of the transformed region $\exp(Y)$, but I'm going for maximum "slick"-factor with this proof as well, so I'd prefer to avoid any calculations more complicated than necessary. In particular, if possible I don't want to use any other properties of the exponential function than those mentioned here.

If there is another entirely different way to prove this nicely, I'm all ears.