How to show Warsaw circle is non-contractible?

The Warsaw circle is defined as a subset of $\mathbb{R}^2$： $$\left\{\left(x,\sin\frac{1}{x}\right): x\in\left(0,\frac{1}{2\pi}\right]\right\}\cup\left\{(0,y):-1\leq y\leq1\right\}\cup C\;,$$ where $C$ is the image of a curve connecting the other two pieces.

A map from Warsaw circle to a single point space seems to be a well-known example showing weak homotopy equivalence is indeed weaker than homotopy equivalence. I am trying to see why the Warsaw circle is non-contractible. It seems intuitively reasonable since two 'ends' of it are connected in some sense, but I failed to give a proof. Any hint would be appreciated. Thank you very much.

Let $W$ denote the Warsaw circle. By collapsing the interval piece down to a point, the quotient space is homeomorophic to $S^1$. This gives a map $f:W\rightarrow W/{\sim} \cong S^1$.

I claim this map is not null homotopic. Believing this for a second, note that for any contractible space $X$ any continuous $f:X\rightarrow Y$ is null homotopic, so this will show that $W$ is not contractible.

So, assume $f$ is null homotopic. Then we can lift it to get a map $\hat{f}:W\rightarrow \mathbb{R}$. Now, $f$ is $1-1$, except on the interval piece. This implies the lift is $1-1$, except, perhaps, on the interval piece. But since the interval piece is connected and it must map into the fiber $\mathbb{Z}$ of the map $\mathbb{R}\rightarrow S^1$, this implies that $\hat{f}$ is 1-1, except that it collapses the interval to a point.

Said another way, $\hat{f}$ descends to an injective map on $W/{\sim}$. Since $W/{\sim}$ is homeomorphic to $S^1$, $\hat{f}$ gives an injective map from $S^1$ to $\mathbb{R}$. But using the intermediate value theorem twice, it's easy to see that there is no injective continuous map from $S^1$ to $\mathbb{R}$.

Its Cech cohomology is equal to $\mathbb Z$ in degree $1$. You can either check this directly, or you could apply Alexander Duality (which relates the reduced Cech cohomology with the reduced homology of the complement.) In this case, since the complement has two path components, the Cech cohomology of the Warsaw circle must be $\mathbb Z$ in degree 1.