= Solution
Write $\tau=x+iy\in\mathfrak h$. For
$$
\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma(1),
$$
the imaginary part of the <Möbius transformation> is
$$
\operatorname{Im}(\gamma\tau)=\frac{y}{|c\tau+d|^2}.
$$
Among the primitive integer pairs $(c,d)$, choose one minimizing the nonzero quantity $|c\tau+d|$. Such a minimum exists because only finitely many <lattice point>[lattice points] lie in a bounded region. Complete $(c,d)$ to a matrix $\gamma\in SL_2(\mathbb Z)$. Then $\gamma\tau$ has maximal imaginary part in its <modular group> orbit.
Applying an integral translation does not change that imaginary part, so arrange
$$
-\frac12\leq\operatorname{Re}(\gamma\tau)\leq\frac12.
$$
If $\operatorname{Im}(\gamma\tau)<\sqrt3/2$, then $|\gamma\tau|<1$. The modular inversion $S:z\mapsto-1/z$ would give
$$
\operatorname{Im}(S\gamma\tau)
=\frac{\operatorname{Im}(\gamma\tau)}{|\gamma\tau|^2}
>\operatorname{Im}(\gamma\tau),
$$
contradicting maximality. Hence every orbit meets the stated region. This is the <reduction to the standard modular region> argument.
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