The standard fundamental domain tessellates the upper half-plane

Every point of $\mathfrak{H}$ lies in some tile, because its orbit meets $\bar{D}$ [F1]; thus ${\bigcup }_{\gamma }\gamma \bar{D}=\mathfrak{H}$. If two tiles have a common interior point, then $\gamma z={\gamma }^{\mathrm{\prime }}{z}^{\mathrm{\prime }}$ with $z,z^{\mathrm{\prime }}\in D$, so $z,z^{\mathrm{\prime }}$ are equivalent points of $D$; by [F1] they are equal and ${\gamma }^{-1}{\gamma }^{\mathrm{\prime }}$ stabilises $z\in D$, which by [F1] has trivial stabiliser, so $\gamma ={\gamma }^{\mathrm{\prime }}$. Hence distinct tiles have disjoint interiors.

$T$ identifies the two vertical sides, and $S$ identifies the two halves of the circular side, fixing $i$. To check incidence, translate one of two meeting tiles to $\bar{D}$. At a boundary point other than $i,\omega ,\omega +1$ the stabiliser is trivial; [F1] then forces the other tile to be $T\bar{D}$, $T^{-1}\bar{D}$, or $S\bar{D}$, according to the side containing that point. Direct substitution shows that these share respectively a full vertical side or the full circular side. Any other tile can meet $\bar{D}$ only at the three exceptional points. Such an intersection has at most one point: each tile is an intersection of three half-planes bounded by vertical lines or circles orthogonal to the real axis, hence is convex along those real-orthogonal circular or vertical geodesics. Indeed, a real Möbius map sending a given geodesic to the imaginary axis carries each bounding half-plane to one whose intersection with that axis is an interval. Two distinct common points would therefore give a common segment, including a nonexceptional point, which is the already listed side case. Thus every nonempty intersection is a side or a vertex, as asserted.

The tiles are invariant by construction. For local finiteness let $K\subset \mathfrak{H}$ be compact and put $a:={\min }_K\mathrm{Im}>0$. If $w=\gamma z\in K$ with $z\in \bar{D}$ and $c\mathrm{\ne }0$, then $a\le \mathrm{Im}w\le 1\mathrm{/}(c^2\mathrm{Im}z)$, so $\mathrm{Im}z\le 1\mathrm{/}a$. Therefore such a tile meets $K$ through the compact set $L:=\bar{D}\cap \{\mathrm{Im}z\le 1\mathrm{/}a\}$; the compact-set finiteness proved in Local charts and the Riemann surface structure of a modular quotient, step 1.1, leaves only finitely many $\gamma$ with $\gamma L\cap K\mathrm{\ne }\mathrm{\varnothing }$. For $c=0$ the maps are translations $T^n$, and the real-part bounds on $K$ and $\mathrm{\mid }\mathrm{Re}z\mathrm{\mid }\le 1\mathrm{/}2$ leave only finitely many $n$. Thus every compact $K$ meets finitely many tiles; a compact disc neighbourhood at each point proves local finiteness.