Your reasoning is spot on, but there are some annoying technical details to heed. I provided a proof in some greater generality.
If $U$ and $V$ are open subsets of a manifold $M$ with $H^1(U)=0$, $H^1(V)=0$ and $U\cap V$ connected, then $H^1(U\cup V)=0$.
It suffices to show that every closed $1$-form on $U\cup V$ is exact. To this end, let $\omega$ be a closed $1$-form on $U\cup V$. Let $\iota_V$ and $\iota_U$ denote the canonical inclusions of $V$ and $U$ into $U\cup V$, respectively. Since the exterior differential commutes with pullback, it follows that $d\iota_{U}^*\omega=\iota_{U}^*d\omega=0$ and, likewise, $d\iota_{V}^*\omega=\iota_{V}^*d\omega=0$. So $\iota_{U}^*d\omega$ and $\iota_{V}^*d\omega$ are closed. But $H^1(V)$ and $H^1(U)$ are trivial, and hence very closed $1$-form on $U$ and $V$, respectively, are exact. That is to say, there exist functions $f_1:U\to \mathbb{R}$ and $f_2:V\to \mathbb{R}$ so that $df_1=\iota_{U}^*\omega$ and $df_2=\iota_{V}^*\omega$. Now, as $U\cap V$ is connected, we have that $f_1\mid_{U\cap V}$ and $f_2\mid_{U\cap V}$ are cohomologous, as $d(f_1-f_2)=df_1-df_2=0$. Since $U\cap V$ is connected, and $d(f_1-f_2)=0$, it follows that $f_1-f_2=c$ for some constant $c$. Thus, the map $F:U\cup V\to \mathbb{R}$ given by
$F(x)=\left\{\begin{array}{ll}f_1(x)&\text{ if }x\in U\\ f_2(x)+c&\text{ if }x\in V\end{array}\right.$
is smooth on $U\cup V$, and $dF=\omega$ by construction. So $\omega$ is exact on $U\cup V$, as desired.
I've been worked on this problem for some time, and i think i probably solved it based on the hint given on the book. Maybe this seems a little long, but it is really not. I tried my best to make this simple. It is really helpful to follow the arguments with a picture for the case $n=2$ in mind.
It is enough to solve this for the closed unit ball $\bar{\mathbb{B}}^n$ in $\mathbb{R}^n$, since any other closed ball $\bar{B}_r(p)$ in $\mathbb{R}^n$ is homeomorphic to $\bar{\mathbb{B}}^n$ by composition of translation $T : \bar{B}_r(p) \to \bar{B}_r(0)$, defined as $x \mapsto x - p$ together with dilation $D : \bar{B}_r(0) \to \bar{\mathbb{B}}^n$, defined as $x \mapsto \frac{x}{r}$.
$\diamond \quad $ As a subspace of $\mathbb{R}^n$, $\bar{\mathbb{B}}^n$ is a second countable Hausdorff space. For any point $p\in \mathbb{B}^n$, the identity map on $\mathbb{B}^n$ serve as the homeomorphism. So we only need to construct homeomorphisms between neighbourhood of points on $\partial \bar{\mathbb{B}}^n=\mathbb{S}^{n-1}$ with open subsets in $\mathbb{H}^n$. To do this we need to consider $\bar{\mathbb{B}}^n$ as a subspace of $\mathbb{R}^{n+1}$.
Consider the stereographic projection from the south pole $\sigma : \mathbb{S}^n\smallsetminus \{S\} \to \mathbb{R}^n$, which is a homeomorphism, defined as
$$
\sigma(x_1,\dots,x_{n+1}) = \frac{(x_1,\dots,x_n)}{1+x_{n+1}}.
$$
For $i=1,\dots,n$, define
$$U_i^{+} =\{ (x_1,\dots,x_n) \in \mathbb{R}^n : x_i > 0 \}, \quad U_i^{-} =\{ (x_1,\dots,x_n) \in \mathbb{R}^n : x_i < 0 \}$$
be $2n$-many open subsets of $\mathbb{R}^n$, and also for $i=1,\dots,n$
$$
\widetilde{U}_i^+ = \{(x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1} : x_i>0 \}, \quad \widetilde{U}_i^- = \{(x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1} : x_i<0 \}
$$
are $2n$-many open subsets in $\mathbb{R}^{n+1}$.
Observe that
$$\sigma^{-1}(U_i^{\pm}) = \mathbb{S}^n \cap \widetilde{U}_i^{\pm}$$
for each $i=1,\dots,n$. That is $\sigma^{-1}$ map $U_i^+$ to the open hemisphere of $\mathbb{S}^n$ where $x_i>0$, and same for $U_i^-$. In particular,
$\bullet \quad \sigma^{-1}$ is an identity map on $$\partial \bar{\mathbb{B}}^n = \mathbb{S}^{n-1} = \{(x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1} : x_{n+1} = 0 \text{ and }\sum_{i=1}^{n} x_i^2 = 1 \},$$
$\bullet \quad $ The inside region, $\mathbb{B}^n \cap U_i^+$, is maped to the upper part ($x_{n+1}>0$) of the hemisphere $\sigma^{-1}(U_i^+)$,
and the outside region, $U_i^+ \smallsetminus \overline{\mathbb{B}^n}$ , is maped to the lower part ($x_{n+1}<0$) of hemisphere $\sigma^{-1}(U_i^+)$.
Since these hemispheres $\sigma^{-1}(U^{\pm}_i)$ homeomorphic to open unit ball $\mathbb{B}^{n}$ via projection map $\pi_i : (x_1,\dots,x_i,\dots,x_{n+1}) \mapsto (x_1,\dots,x_{i-1},x_{i+1},\dots,x_{n+1})$, then by restricting the composition map $\pi_i \circ \sigma^{-1} : \mathbb{R}^n \to \mathbb{R}^n$ to $U_i^{\pm} \cap \bar{\mathbb{B}}^n$, we obtain the desired homeomorphisms
$$
\varphi:=\pi_i \circ (\sigma^{-1})|_{U_i^{\pm}\cap \bar{\mathbb{B}}^n} : U_i^{\pm}\cap \bar{\mathbb{B}}^n \to \mathbb{H}^n,
$$
with domains cover $\partial\bar{\mathbb{B}}^n = \mathbb{S}^{n-1}$. By construction, any $p\in \mathbb{S}^{n-1}$ must contained in one such neighbourhoods, with $\varphi(p) \in \partial \mathbb{H}^n$ and $\varphi(U_i^{\pm}\cap \bar{\mathbb{B}}^n)$ is an open half unit ball in $\mathbb{H}^n$. Therefore, $\bar{\mathbb{B}}^n$ is an $n$-manifold with boundary with manifold boundary is equal to its topological boundary $\mathbb{S}^{n-1}$.
Note that by similar way we can show that the complement of any open ball is an $n$-manifold with boundary, with its topological boundary as the manifold boundary. Only this time we use stereographic projection form the north pole.
Best Answer
The Stereographic projection is a diffeomorphism $f:\mathbb{S}∖\{p\}→\mathbb{R}^n$, with $f(q)=0∈\mathbb{R}^n$. So the restriction $\mathbb{S}^n∖\{p,q\}→\mathbb{R}^n∖\{0\}$ is the desired diffeomorphism.