We define $\Omega :=\left\lbrace z \in \mathbb{H}\colon -\frac{1}{2} \leq \operatorname{Re}z \leq \frac{1}{2} \wedge |z| \geq 1\right\rbrace$.
I want to show that the following holds:
$$ \forall \, z \in \Omega^{\circ} \colon (\text{SL}_2(\mathbb{Z}) \circ z) \cap \Omega = \left\lbrace z \right\rbrace. \ \ \ \ \ \ (\star)$$
I managed to show that for $w \in \mathbb{H}$ we have that $(\text{SL}_2(\mathbb{Z}) \circ w) \cap \Omega \neq \varnothing$ and now I want to prove $(\star)$ for $z \in \Omega^{\circ}$.
Suppose $z\in \Omega^\circ$ with $\gamma z\in \Omega^\circ$ for $\gamma\not=\pm 1$ in $SL_2({\mathbb{Z}})$. By taking $\gamma \to \gamma^{-1}$, assume without loss of generality that $\text{Im}(z) \leq \text{Im}(\gamma z)$. Clearly $\gamma = \begin{pmatrix}a & b \\c & d\end{pmatrix}$ with $c\not= 0$. Then $$\text{Im} (\gamma z) = \frac{\text{Im}(z)}{\left|cz+d\right|^2}\leq \frac{{\text{Im}(z)}}{\text{Im}(cz+d)^2} = \frac{1}{c^2\, \text{Im}(z)}.$$ But every $w\in \Omega$ has $\text{Im}(w)^2 = |w|^2 - \text{Re}(w)^2 \geq \frac{3}{4}$, so $$c^2 \leq \frac{4}{3}.$$ Hence $c = \pm 1$. But then $|cz + d| \geq |z| > 1$ (since $d\in {\mathbb{Z}}$), and so $$\text{Im}(\gamma z) = \frac{\text{Im}(z)}{|cz + d|^2} < \text{Im}(z).$$ The result follows from this contradiction.