The closed disk $D^2$ is a connected Hausdorff topological $2$-manifold with boundary

The ambient plane and its subspaces. By [F1] the metric topology of $\mathbb{C}$ is the Euclidean topology of ${\mathbb{R}}^2$ under $x+iy\mapsto (x,y)$; ${\mathbb{R}}^2$ is Hausdorff and second countable with the countable basis of rational open boxes. Since $D^2\subseteq \mathbb{C}$ carries the subspace topology, [F2] makes $D^2$ Hausdorff and second countable.

$D^2$ is nonempty, convex and connected. Clearly $0\in D^2$. If $z,w\in D^2$ and $t\in [0,1]$, then $\mathrm{\mid }(1-t)z+tw\mathrm{\mid }\le (1-t)\mathrm{\mid }z\mathrm{\mid }+t\mathrm{\mid }w\mathrm{\mid }\le 1$ by multiplicativity and subadditivity of the modulus in [F3], so $D^2$ is convex; it is a nonempty convex subset of ${\mathbb{R}}^2$ in the sense of [F4], hence contractible, hence path-connected, hence connected.

Interior charts. Let $z\in \mathbb{C}$ with $\mathrm{\mid }z\mathrm{\mid }<1$ and put $r:=\frac{1}{2}(1-\mathrm{\mid }z\mathrm{\mid })>0$. If $\mathrm{\mid }w-z\mathrm{\mid }<r$ then $\mathrm{\mid }w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w-z\mathrm{\mid }<\mathrm{\mid }z\mathrm{\mid }+r<1$ by [F3], so $B(z,r)\subseteq D^2$ and $U:=D^2\cap B(z,r)=B(z,r)$ is an open neighbourhood of $z$ in $D^2$ that is open in $\mathbb{C}$ as well. The translation $\psi (w):=w+2i$ is continuous with continuous inverse $u\mapsto u-2i$ by [F3], hence a homeomorphism of $\mathbb{C}$; its restriction to $U$ is therefore a homeomorphism of $U$ onto the open set $\psi (U)\subseteq {\mathbb{R}}^2$, and for $w\in U$ one has $\mathrm{Im}w\ge -\mathrm{\mid }w\mathrm{\mid }>-1$ and hence $\mathrm{Im}\psi (w)=\mathrm{Im}w+2>1>0$, so $\psi (U)$ lies in the open upper half-plane and is in particular a relatively open subset of ${\mathbb{H}}^2$ containing $\psi (z)$.

The two-sided inverse of the boundary formula. Let $q\in \mathbb{C}$ with $\mathrm{\mid }q\mathrm{\mid }=1$, put ${\phi }_q(w):=i(q-w)\mathrm{/}(q+w)$ for $w\mathrm{\ne }-q$, and put $\psi (u):=q(i-u)\mathrm{/}(i+u)$ for $u\mathrm{\ne }-i$. Both are defined on the sets where they are used below, because $\mathrm{\mid }q-(-q)\mathrm{\mid }=2>1$ gives $-q\mathrm{\notin }{V}_q$ and because $\mathrm{\mid }i+u\mathrm{\mid }\ge \mathrm{Im}u+1\ge 1>0$ whenever $\mathrm{Im}u\ge 0$. For $u\mathrm{\ne }-i$, $\frac{i(q-\psi (u))}{q+\psi (u)}=\frac{i(1-\frac{i-u}{i+u})}{1+\frac{i-u}{i+u}}=\frac{i\frac{(i+u)-(i-u)}{i+u}}{\frac{(i+u)+(i-u)}{i+u}}=\frac{i\cdot 2u}{2i}=u,$ so ${\phi }_q\circ \psi =\mathrm{id}$ on $\mathbb{C}\setminus \{-i\}$, in particular on $\{\mathrm{Im}u\ge 0\}$, and for $w\mathrm{\ne }-q$, $\psi ({\phi }_q(w))=\frac{q(i-\frac{i(q-w)}{q+w})}{i+\frac{i(q-w)}{q+w}}=\frac{q\frac{i2w}{q+w}}{\frac{i2q}{q+w}}=w,$ so $\psi \circ {\phi }_q=\mathrm{id}$ on $\mathbb{C}\setminus \{-q\}$, which contains $V_q$. Hence ${\phi }_q$ and $\psi$ are mutually inverse bijections between $\mathbb{C}\setminus \{-q\}$ and $\mathbb{C}\setminus \{-i\}$, and in particular ${\phi }_q$ is injective on $D^2\setminus \{-q\}$.

Which points of the plane are carried into the half-space. Let $w\in D^2\setminus \{-q\}$ and multiply numerator and denominator of ${\phi }_q(w)$ by $\bar{q}+\bar{w}$, which is the conjugate of $q+w$: using $q\bar{q}=\mathrm{\mid }q{\mathrm{\mid }}^2=1$, $w\bar{w}=\mathrm{\mid }w{\mathrm{\mid }}^2$ and $s-\bar{s}=2i\mathrm{Im}s$ for $s:=q\bar{w}$ gives ${\phi }_q(w)=\frac{i(q-w)(\bar{q}+\bar{w})}{\mathrm{\mid }q+w{\mathrm{\mid }}^2}=\frac{i(\mathrm{\mid }q{\mathrm{\mid }}^2+q\bar{w}-w\bar{q}-\mathrm{\mid }w{\mathrm{\mid }}^2)}{\mathrm{\mid }q+w{\mathrm{\mid }}^2}=\frac{i(1-\mathrm{\mid }w{\mathrm{\mid }}^2+2i\mathrm{Im}(q\bar{w}))}{\mathrm{\mid }q+w{\mathrm{\mid }}^2},$ so $\mathrm{Im}{\phi }_q(w)=(1-\mathrm{\mid }w{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }q+w{\mathrm{\mid }}^2$, which is $\ge 0$ exactly when $\mathrm{\mid }w\mathrm{\mid }\le 1$: thus ${\phi }_q$ maps $D^2\setminus \{-q\}$ into the closed upper half-plane $\{\mathrm{Im}u\ge 0\}$. Conversely, if $\mathrm{Im}u\ge 0$ then $\mathrm{\mid }\psi (u){\mathrm{\mid }}^2=\frac{\mathrm{\mid }i-u{\mathrm{\mid }}^2}{\mathrm{\mid }i+u{\mathrm{\mid }}^2}=\frac{(\mathrm{Re}u{)}^2+(1-\mathrm{Im}u{)}^2}{(\mathrm{Re}u{)}^2+(1+\mathrm{Im}u{)}^2}\le 1,$ because the last inequality is equivalent to $(1-\mathrm{Im}u{)}^2\le (1+\mathrm{Im}u{)}^2$, that is to $-2\mathrm{Im}u\le 2\mathrm{Im}u$. Hence $\psi$ maps $\{\mathrm{Im}u\ge 0\}$ into $D^2$. Since ${\phi }_q\circ \psi$ is the identity by step 1.4, ${\phi }_q$ is surjective onto $\{\mathrm{Im}u\ge 0\}$ and $\psi$ is injective; by step 1.4, ${\phi }_q$ is also injective. So ${\phi }_q:D^2\setminus \{-q\}\to \{\mathrm{Im}u\ge 0\}$ is a bijection with inverse $\psi$.

${\phi }_q$ and $\psi$ are continuous, hence homeomorphisms. For $w,w_0\in D^2\setminus \{-q\}$ expansion gives ${\phi }_q(w)-{\phi }_q(w_0)=\frac{i(q-w)(q+w_0)-i(q-w_0)(q+w)}{(q+w)(q+w_0)}=\frac{2iq(w_0-w)}{(q+w)(q+w_0)},$ so by multiplicativity of the modulus and $\mathrm{\mid }i\mathrm{\mid }=\mathrm{\mid }q\mathrm{\mid }=1$, $\mathrm{\mid }{\phi }_q(w)-{\phi }_q(w_0)\mathrm{\mid }=\frac{2\mathrm{\mid }w-w_0\mathrm{\mid }}{\mathrm{\mid }q+w\mathrm{\mid }\mathrm{\mid }q+w_0\mathrm{\mid }}\mathrm{.}$ Let $\delta :=\mathrm{\mid }q+w_0\mathrm{\mid }>0$ and suppose $\mathrm{\mid }w-w_0\mathrm{\mid }\le \delta \mathrm{/}2$; then $\mathrm{\mid }q+w\mathrm{\mid }\ge \mathrm{\mid }q+w_0\mathrm{\mid }-\mathrm{\mid }w-w_0\mathrm{\mid }\ge \delta \mathrm{/}2$ by subadditivity, so $\mathrm{\mid }{\phi }_q(w)-{\phi }_q(w_0)\mathrm{\mid }\le 4\mathrm{\mid }w-w_0\mathrm{\mid }\mathrm{/}{\delta }^2$, which is $<\epsilon$ as soon as $\mathrm{\mid }w-w_0\mathrm{\mid }<\min (\delta \mathrm{/}2,\epsilon {\delta }^2\mathrm{/}4)$. This is the $\epsilon$-$\delta$ condition of [F5] for continuity of ${\phi }_q$ at $w_0$. The same expansion with $i$ in place of $q$ and $u$'s in place of $w$'s gives $\mathrm{\mid }\psi (u)-\psi (u_0)\mathrm{\mid }=2\mathrm{\mid }u-u_0\mathrm{\mid }\mathrm{/}(\mathrm{\mid }i+u\mathrm{\mid }\mathrm{\mid }i+u_0\mathrm{\mid })$, and $\mathrm{\mid }i+u_0\mathrm{\mid }\ge 1$ for $\mathrm{Im}{u}_0\ge 0$, so $\psi$ is continuous on the closed upper half-plane as well. By step 2.1 and [F5], ${\phi }_q$ is a homeomorphism of $D^2\setminus \{-q\}$ onto $\{\mathrm{Im}u\ge 0\}$; consequently ${\phi }_q(V_q)$ is a relatively open subset of ${\mathbb{H}}^2$ by [F6], because $V_q$ is open in $D^2$ and hence in $D^2\setminus \{-q\}$, and it contains ${\phi }_q(q)=0$.

Conclusion. Step 3.1 shows that for $\mathrm{\mid }q\mathrm{\mid }=1$ the restriction of ${\phi }_q$ to the neighbourhood $V_q$ of $q$ in $D^2$ is a homeomorphism onto a relatively open subset of ${\mathbb{H}}^2$ containing $0$, and step 1.3 provides the corresponding chart at every point with $\mathrm{\mid }z\mathrm{\mid }<1$. Step 1.2 shows $D^2$ is nonempty and connected and step 1.1 shows it is Hausdorff and second countable, so every point of $D^2$ has a neighbourhood homeomorphic to a relatively open subset of ${\mathbb{H}}^2$: by [F6], $D^2$ is a topological $2$-manifold with boundary, as claimed.