quasi-circle has trivial homology group

Question

As I am reviewing my algebraic topology material, I got confused by something seems really basic.

For a quasi-circle $Q$ formed by $y=\sin(1/x)$ with $x \in [0,1]$, $[-1,1]$ on y-axis, and an arc $c$ connecting the $(0,0)$ and $(1,0)$ (as in Hatcher's book 1.3.7), I want to prove that it has trivial homology groups.

I have two ways to do it, which seems to give me different results.

Let $A$ contain the arc $c$, the verticle $[-1,1]$, and a bit of the topologist's sine curve on both ends. Let $B$ contain the topologist's sine curve (not including the verticle part). Clearly, both $A$ and $B$ are contractible, thus have trivial homogology groups.

In this case, we get $\tilde{H}_n(A) \oplus \tilde{H}_n(B) = 0$ for all $n$, which forces $\tilde{H}_n(Q) \cong \tilde{H}_n(A \cap B)$. In this case, $\tilde{H}_n(A \cap B) = 0$ for all $n$ except for $\tilde{H}_0(A \cap B) \cong \mathbb{Z}$, as it has two connected components.

However, I can also choose $A$ such that it only contains some topologist's sine curve on the 1 side but not on the 0 side. In which case I would think their interior also covers the quaso circle, and we would have all homology groups are trivial.

It seems I made a simple mistake somewhere, can anyone point it out? Is there an easier way to find the homology group than I did?

$\textbf{Edit}$ After seeing the comment, I realized that when $A$ contains both the vertical line segment and the topologist sine curve, it can not be contracted. So I thought let $A$ contain the verticle segment and the arc, and let $B$ contain the topologist sine curve and the arc. That would give result same as the second approach above.

Answer 1

Claim: for any singular simplex $\sigma:\Delta^n \to Q$, there exists some $\delta>0$ such that $\sigma(\Delta^n) \cap ((0,\delta) \times \mathbb{R})$ is empty. Given this fact, any given simplex lies in a contractible subspace of $Q$. Since homotopic maps are homologous, this gives that every singular chain (and so every singular cycle) is nullhomologous. In other words $\tilde H_n(Q)=0$ for all $n$.

Proof of claim: if the claim were not true, then there would be a sequence of points $p_k \in \Delta^n$ such that the $x$-coordinates of $\sigma(p_k)$ approach $0$ monotonically. Since the $n$-simplex is compact, this limit is achieved. I.e. we have $p_k \to p_*$ in $\Delta^n$ with $\sigma(p_*) \in \{0\} \times [-1,1]$. But $\Delta^n$ is path connected, so there is a path $p_1 \rightsquigarrow p_2 \rightsquigarrow \dotsb \rightsquigarrow p_*$ in $\Delta^n$ whose image in $Q$ connects a point on the sine curve with the vertical segment. (We can go far enough out in the path to make sure it's actually "jumping the gap" rather than going round the bottom arc of the quasi circle.) This is impossible.