Prove there are two points making two regions not conformally equivalent Show that there exist distinct points $z_0,z_1$ in the open unit disk $D$ so that $D - \{1/2, 1/3\}$ and $ D - \{z_0, z_1\}$ are not conformally equivalent. 
Thoughts so far: I'm not sure where to go with this, so a hint to get me started would be appreciated. 
 A: Call $A$ and $B$ the two open sets. Assume that a biholomorphism $\phi:A\to B$ exists and call $\psi:B\to A$ its inverse.
$\phi$ and $\psi$ extend to holomorphic maps $\overline{\phi},\overline{\psi}:D\to D$ (why?) and
$\overline{\phi}\circ\overline{\psi}=\overline{\psi}\circ\overline{\phi}=id$ (why?). So $\overline{\phi}:D\to D$ is an automorphism of the disk and thus it maps $\{1/2, 1/3\}$ to $\{z_0,z_1\}$.
Now choose any $z_0\in D$ (you can choose for instance $z_0=1/2$ but it's not important). We will find a suitable $z_1$ s.t. such a $\overline{\phi}$ cannot exist. We split the problem in two cases:
If $\overline{\phi}(1/2)=z_0$: choose two automorphisms of $D$, $\alpha$ and $\beta$, s.t. $\alpha(0)=1/2$ and $\beta(z_0)=0$. Now $\beta\circ\overline{\phi}\circ\alpha(0)=0$ and $\beta\circ\overline{\phi}\circ\alpha$ is an automorphism, so by Schwarz lemma $|\beta\circ\overline{\phi}(1/3)|=|\alpha^{-1}(1/3)|$ (why?), i.e. $|\beta(z_1)|=|\alpha^{-1}(1/3)|$, i.e. $z_1$ belongs to $\beta^{-1}(S_1)$, where $S_1$ is the circle centered at $0$ with radius $|\alpha^{-1}(1/3)|$.
The other case, where $\overline{\phi}(1/3)=z_0$, is completely analogous and gives $z_1\in (\beta')^{-1}(S_2)$ (now $S_2$ has radius $|(\alpha')^{-1}(1/2)|$) for suitable automorphisms $\alpha',\beta'$. So in any case, if $\phi$ exists, we have $z_1\in\beta^{-1}(S_1)\cup(\beta')^{-1}(S_2)=:K$, which is a compact set. Thus $K\subsetneq D$. Conclude.
