This isn't the inductive proof you were looking for, but hopefully it's illuminating.
First recall the following theorem about Čech resolutions for locally finite closed covers:
Theorem [Godement, II, Thm. 5.2.1]. Let $X$ be a topological space, $\mathscr{F}$ a sheaf of abelian groups on $X$, and $\mathfrak{M} = (M_i)_{ \in I}$ a locally finite closed cover of $X$. Then, the complex $\mathscr{C}^\bullet(\mathfrak{M},\mathscr{F})$ as defined in [Hartshorne, p. 220] is a resolution for $\mathscr{F}$.
In our case, write $Y = \bigcup_{i=0}^n H_i$ where the $H_i$ are hyperplanes in general position; this is a finite closed cover of $Y$. Letting $\mathscr{F} = \mathbf{Z}\rvert_Y$ in the theorem above, and then pushing forward along the inclusion $i\colon Y \to X$, we have a resolution
$$
0 \longrightarrow \mathbf{Z}_Y \longrightarrow \bigoplus_i \mathbf{Z}_{H_i} \longrightarrow \bigoplus_{i<j} \mathbf{Z}_{H_i \cap H_j} \longrightarrow \cdots \longrightarrow \bigoplus_{i_1<i_2<\cdots<i_n} \mathbf{Z}_{H_{i_1} \cap H_{i_2} \cap \cdots \cap H_{i_n}} \longrightarrow 0
$$
of $\mathbf{Z}_Y$ on $X$, since $\bigcap_{i=0}^n H_i = \emptyset$ by the hypothesis that the $H_i$ are in general position. This is also a flasque resolution since each intersection $H_{i_1} \cap H_{i_2} \cap \cdots \cap H_{i_\ell}$ is irreducible, and since direct images of flasque sheaves are flasque.
Now note that after taking global sections in the resolution above, we obtain the chain complex for the simplicial homology of $S^{n-1}$ (here we use that the hyperplanes intersect in the "expected" way, and so the hyperplanes must be chosen generally). This implies $H^{n-1}(X,\mathbf{Z}_Y) = H_0(S^{n-1},\mathbf{Z}) = \mathbf{Z}$ if $n > 1$ and $\mathbf{Z}^2$ if $n = 1$. Now using the long exact sequence on cohomology from your short exact sequence
$$0 \longrightarrow \mathbf{Z}_U \longrightarrow \mathbf{Z} \longrightarrow \mathbf{Z}_Y \longrightarrow 0,$$
we have an exact sequence
$$\cdots \longrightarrow H^{n-1}(X,\mathbf{Z}) \longrightarrow H^{n-1}(Y,\mathbf{Z}) \longrightarrow H^n(X,\mathbf{Z}_U) \longrightarrow H^n(X,\mathbf{Z}) \longrightarrow \cdots.$$
Finally, if $n > 1$, then $H^{n-1}(X,\mathbf{Z}) = H^n(X,\mathbf{Z}) = 0$ since $\mathbf{Z}$ is flasque on $X$, and so we have that $H^n(X,\mathbf{Z}_U) = \mathbf{Z}$. If $n = 1$, then the exact sequence is
$$
0 \longrightarrow \mathbf{Z} \longrightarrow \mathbf{Z}^2 \longrightarrow H^1(X,\mathbf{Z}_U) \longrightarrow 0
$$
and so and $H^1(X,\mathbf{Z}_U) = \mathbf{Z}$ if $n = 1$ as well.
Since I wrote up a proof of the Theorem I cited above, I thought I might as well put it here for completeness, especially since the only English reference I could find assumes $X$ is $T_3$. First, we show the following analogue of the glueing axiom for locally finite closed covers.
Lemma [Godement, II, Thm. 1.3.1]. Let $\mathscr{F}$ be a sheaf on $X$ and suppose $\mathfrak{M}$ is a locally finite closed cover, and consider $\mathscr{F}$ as its espace étalé $\operatorname{Sp\acute{e}}(\mathscr{F}) \to X$. Suppose we are given continuous maps $s_i \colon M_i \to \operatorname{Sp\acute{e}}(\mathscr{F})$ such that $s_i\rvert_{M_i \cap M_j} = s_j\rvert_{M_i \cap M_j}$ for each $i,j$. Then, there exists a section $s\colon X \to \operatorname{Sp\acute{e}}(\mathscr{F})$ such that $s\rvert_{M_i} = s_i$ for all $i \in I$.
Proof. It is clear that the $s_i$ glue to give a (unique) section $s\colon X \to \operatorname{Sp\acute{e}}(\mathscr{F})$ such that $s\rvert_{M_i} = s_i$; it suffices to show this section is continuous. For any point $x \in X$, since the cover $\mathfrak{M}$ is locally finite, there is an open set $U(x)$ that intersects only finitely many $M_i$; call these $M_{i_1},\ldots,M_{i_n}$. Now on $U(x)$, the $s_{i_1},\ldots,s_{i_n}$ glue to form a section $t \colon U(x) \to \operatorname{Sp\acute{e}}(\mathscr{F})$ by, e.g., [Munkres, Thm. 18.3]; since $s = t$ on $U(x)$, we see that $s$ is continuous since continuity is local. $\blacksquare$
Now we can prove the Theorem.
Proof of Theorem. We define $\epsilon \colon \mathscr{F} \to \mathscr{C}^0$ by taking the product of the natural maps $\mathscr{F} \to f_*(\mathscr{F}\rvert_{M_i})$ for $i \in I$. The map $\epsilon$ is injective: every point $x \in X$ lies in some $M_i$, there the map on stalks $\mathscr{F}_x \to (f_*(\mathscr{F}\rvert_{M_i}))_x$ is an isomorphism, and so $\epsilon$ is injective. Exactness at $\mathscr{C}^0$ follows from the Lemma. It therefore suffices to show exactness at $\mathscr{C}^n$ for each $n \ge 1$; it suffices to show exactness on stalks.
Let $x \in X$. For each $n \ge 1$, we define a map
$$
k \colon \mathscr{C}^n(\mathfrak{M},\mathscr{F})_x \longrightarrow \mathscr{C}^{n-1}(\mathfrak{M},\mathscr{F})_x
$$
as follows. Given $\alpha_x \in \mathscr{C}^{n-1}(\mathfrak{M},\mathscr{F})_x$, it is represented by a section $\alpha \in \Gamma(U,\mathscr{C}^n(\mathfrak{M},\mathscr{F}))$ over a neighborhood $U$ of $x$; since $\mathfrak{M}$ is a locally finite cover, we can assume that $U$ intersects only finitely many $M_{j_1},\ldots,M_{j_n}$, and since they are closed, we can assume they all contain $x$ after possibly subtracting some $M_{j_\ell}$ off from $U(x)$. By replacing $\mathfrak{M}$ with $\mathfrak{M} \cap U$, we can assume $\mathfrak{M}$ is in fact finite. Now let $j \in \{j_1,\ldots,j_n\}$ be arbitrary; for any $p$-tuple $i_0 < \cdots < i_{p-1}$, we set
$$
(\beta_{i_0,\ldots,i_{p-1}})_x = (\alpha_{j,i_0,\ldots,i_{p-1}})_x
$$
using the notational convention of Rem. III.4.0.1. Since there are only finitely many indexes $j$, we see by the Lemma that these $\beta$ glue to give a section $\beta_{i_0,\ldots,i_{p-1}} \in \mathscr{F}\rvert_{M_{i_0,\ldots,i_{p-1}}}(M_{i_0,\ldots,i_{p-1}})$ and hence we get an element $\beta \in \Gamma(U,\mathscr{C}^n(\mathfrak{M} \cap U,\mathscr{F}))$. Now we define $k$ above as $\alpha_x \mapsto \beta_x$. By the same argument as in Lem. III.4.2, $k$ is a homotopy operator for the complex $\mathscr{C}^\bullet_x$, that is, we have $(dk + kd)(\alpha_x) = \alpha_x$ for any germ $\alpha_x \in \mathscr{C}^p_x$, and so the identity map is homotopic to the zero map. Finally, it follows that the cohomology groups $h^p(\mathscr{C}^\bullet_x)$ of this complex are $0$ for $p \ge 1$. $\blacksquare$
For (a), your resolution is not right, because $(\prod_{p\in S^1}i_p(\mathbb{Q}))/\mathbb{Z}$ is not $\prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$. Or in other words, the kernel of $\prod_{p\in S^1}i_p(\mathbb{Q})\rightarrow \prod_{p\in S^1}i_p(\mathbb{Q}/\mathbb{Z})$ is not $\mathbb{Z}$ (this is $\prod_{p\in S^1}i_p(\mathbb{Z})$).
For (b), you cannot mimick example 4.0.4 to compute the Cech cohomology. In fact, the higher Cech cohomology of this sheaf vanishes. Let $U,V$ your cover (which is good), and consider the Cech complex
$$ R(U)\times R(V)\rightarrow R(U\cap V)$$
given by $(f,g)\mapsto f_{|U\cap V}-g_{|U\cap V}$.
The kernel of this map are couple of functions $(f,g)$ that agree on the intersection, and hence patch together to form a unique function on $S^1$. So $\overset{\vee}{H^0}(\{U,V\},R)=R(S^1)$ as expected.
But I claim that the map is onto, so that $\overset{\vee}{H^1}(\{U,V\},R)=0$. Indeed, let $f\in R(U\cap V)$, also let $u,v$ be two functions on $S^1$ such that $u$ has (compact) support in $U$, $v$ has (compact) support in $V$ and $u+v=1$. In particular the function $u$ is zero in a neighborhood of $S^1\setminus V$ and $v$ is zero in a neighborhood of $S^1\setminus U$. The function $uf$ can then be extended to $U$ because it is zero in a neighborhood of $U\setminus U\cap V$. Similarily, the function $vf$ can be extended to $V$. And $(uf,-vf)\mapsto uf+vf=f$ so that the map is onto.
Here $u,v$ are called a partition of unity. A sheaf like $R$ with partitions of unity is called a fine sheaf. This argument (of a similar one with more than two open sets in the cover) shows that the higher Cech cohomology groups of a fine sheaf over a paracompact space vanish. On a paracompact space, Cech and derived functor cohomology agree so the cohomology of a fine sheaf is trivial.
The difference with the example 4.0.4 is that there is no partition of unity in a constant sheaf.
Best Answer
I think the issue is that the answer to your question in the title is negative, since it's irreducibility that you're really interested in to argue that this sheaf is flasque. Explicitly, if you take the union of the $x$ and $y$ axes in the affine plane $Y=V(xy) \subseteq \mathbf{A}^2_k$ and the constant sheaf $\underline{\mathbf{Z}} \in \text{Sh}_{\text{Ab}}(Y)$ (which is the restriction of the constant sheaf $\underline{\mathbf{Z}}$ on $\mathbf{A}^2_k$ to $X$) then the global sections are given by $\Gamma(Y,\underline{\mathbf{Z}}) = \mathbf{Z}$ since $Y$ is connected, but $Y$ has the disconnected open subset $U = D(x) \cup D(y) \subseteq Y$. You can see $U$ is disconnected by the equality $$ D(x) \cap D(y) = D(xy) = \emptyset $$ or by just checking that $U$'s global sections are given by the equaliser $$ \text{eq}(k[x,x^{-1}] \times k[y,y^{-1}] \rightrightarrows (0)) = k[x,x^{-1}] \times k[y,y^{-1}] $$ which has precisely two non-trivial idempotents - this implies $\Gamma(U,\underline{\mathbf{Z}}) = \mathbf{Z} \oplus \mathbf{Z}$ and thus the restriction map $$ \Gamma(Y,\underline{\mathbf{Z}}) \to \Gamma(U,\underline{\mathbf{Z}}) $$ can't be surjective.
Hope this helps :)