= Solution
Fix $z_0\in\mathcal U$. We prove the <Riemann mapping theorem> in this case by maximizing a normalized <derivative>. Choose $a\notin\mathcal U$. Since $\mathcal U$ is a <simply connected domain> and $z-a$ never vanishes, it has a <holomorphic square root> $g$. This $g$ is <injective>, and $g(\mathcal U)$ is disjoint from $-g(\mathcal U)$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal U)$ and $r>0$ with $B(b,r)\subset g(\mathcal U)$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded <univalent function>. Scaling and composing with an <automorphism of the unit disk> produces an <injective> <holomorphic function> $\phi:\mathcal U\to\mathbb D$ with $\phi(z_0)=0$ and $\phi'(z_0)>0$.
Let $\mathcal F$ be the family of all such normalized <univalent functions>. A <Cauchy estimate> in a small disk about $z_0$ bounds their <derivatives>, so $M=\sup_{\phi\in\mathcal F}\phi'(z_0)$ is finite and positive. Choose a maximizing sequence in this <normal family>. <Montel theorem> gives a subsequence converging uniformly on <compact> subsets to $\phi$. Its <derivative> at $z_0$ is $M>0$. The <maximum modulus principle> puts its image in $\mathbb D$, and <Hurwitz's theorem> implies that a nonconstant limit of <injective> <holomorphic functions> is <injective>. Thus the maximum is attained.
If $a_1\in\mathbb D$ is omitted, then $a_1\ne0$. Put $T_a(w)=(w-a)/(1-\overline a w)$. A <holomorphic square root> $v$ of $T_{a_1}\circ\phi$ exists, is <injective>, and takes values in $\mathbb D$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in(0,1)$, the new <derivative> has magnitude
$$
M\frac{1-s^2}{2\sqrt s(1-s)}
=M\frac{1+s}{2\sqrt s}>M,
$$
a contradiction. Therefore $\phi$ maps onto $\mathbb D$. The <Cayley transform between the half-plane and disk> now gives
$$
\boxed{f(z)=i\,\frac{1+\phi(z)}{1-\phi(z)}:\mathcal U\xrightarrow{\sim}\mathbb H,\qquad
\operatorname{Im}f(z)=\frac{1-|\phi(z)|^2}{|1-\phi(z)|^2}>0.}
$$
Its inverse is <holomorphic> by the <holomorphic inverse function theorem>, so this is a <biholomorphism>. The proper-subset hypothesis was used to choose $a\notin\mathcal U$; the whole <complex plane> cannot be mapped this way, by <Liouville theorem>.
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