The standard stem system cuts the punctured disk open to a disk

A concrete model of a slit tip. Choose pairwise disjoint small round disks $B_i$ about the punctures, meeting only their own stems. In polar coordinates about $q_i$, put the stem radius at angle $0$. The cut of $B_i\setminus \{q_i\}$ is $(0,R_i]\times [0,2\pi ]$, not a compact annulus or disk. Adjoin its missing tip by forming $E_i=([0,R_i]\times [0,2\pi ])\mathrm{/}(\{0\}\times [0,2\pi ])$: the whole zero-radius edge is collapsed to one point. This is a closed disk (a rectangle with one edge collapsed, equivalently a triangle), whose boundary is the outer circular arc and the two radial sides joined at the tip. The map $(r,\theta )\mapsto q_i+r{e}^{\mathrm{i}\theta }$ extends continuously to the tip and, on identifying the two radial sides, gives the filled disk $B_i$. Deleting the tip before regluing gives $B_i\setminus \{q_i\}$.

The outer piece. Put $\mathrm{\Omega }=D^2\setminus {\bigcup }_i\mathrm{int}{B}_i$. Open it along the truncated stems from $d$ to $C_i=\mathrm{\partial }{B}_i$, separating all sectors at their common endpoint $d$. To see that the result is a disk, first thicken each truncated stem to a narrow strip, with strips disjoint away from a small half-disk at $d$. The region left between these strips and the $B_i$ has one polygonal Jordan boundary: tracing it visits the outer boundary once and makes one detour along both sides of each strip and around its associated hole. It is a closed disk by [F1]. Shrinking the strip widths gives the same cut topology, since each strip-side collar has a rectangular coordinate chart and changing its width is a homeomorphism. Each opened $C_i$ is a closed boundary arc $A_i$; at $d$ there are $n+1$ sector copies when $n>0$, rather than just two copies for the whole star. For $n=0$ the outer piece is $D^2$.

Attaching the completed tips. Glue the circular boundary arc of $E_i$ to $A_i$ with matching radial-side endpoints. Gluing two disks along a proper closed boundary arc gives a disk: map the two disks to the upper and lower half-disks, with the glued arcs as their common diameter, and use these maps on the quotient. Applying this construction finitely many times to the outer disk and the $E_i$ yields a compact disk $H$. Its boundary consists of the outer boundary arc and the two sides of every stem, with each pair joined at $q_i$ and with consecutive sides joined at the appropriate sector copy of $d$.

The quotient and its topology. Identify matching points on each pair of stem sides, including the copies of $d$. The quotient map $\pi :H\to D^2$ is the ordinary coordinate map off the cuts and the polar map of step 1.1 at a tip. It is a continuous surjection whose fibers are exactly these prescribed identifications; compactness of $H$ and the Hausdorff property of $D^2$ show that its quotient is the filled disk $D^2$. Removing the images $q_i$ of the added tips recovers $X=D^2\setminus Q_n$. Thus the compact disk assertion concerns the completed cut, with both filled and punctured quotients accounted for.

Lifting maps. A homeomorphism $f$ as in the statement preserves each local side of every stem: an orientation-reversing map would reverse the outer boundary, which $f$ fixes pointwise, and an orientation-preserving map fixing an oriented stem cannot interchange its sides. Thus it lifts on the open cut surface, fixing both stem-side copies and all sector copies of $d$. Near a tip, continuity of $f$ at $q_i$ implies that points with radius tending to zero have image radius tending to zero, uniformly in their angle; the collapsed-edge model therefore extends the lift continuously by fixing the tip. The same argument applies to $f^{-1}$, so the lift is a boundary-fixed homeomorphism of $H$.

Regluing maps and isotopies. A boundary-fixed homeomorphism $g$ of $H$ respects every fiber of $\pi$ and induces a homeomorphism of the filled disk, with inverse induced by $g^{-1}$. It fixes the outer boundary, stems and punctures pointwise. If $G:H\times I\to H$ is a boundary-fixed isotopy, the map $(z,t)\mapsto \pi (G(z,t))$ is constant on the fibers of $\pi \times {\mathrm{id}}_I$; this map is a quotient map because its domain is compact and its target is Hausdorff. It therefore induces a continuous isotopy $D^2\times I\to D^2$, including at every puncture uniformly in time. This proves the asserted correspondence and the isotopy version needed by consumers.