Prove that the "theta-space" is not homeomorphic to $S^{1}$. Let $X$ be the "$\theta$-space":
\begin{equation*}
X = \{ (x,y) \in \mathbb{R}^{2} \colon x^{2} + y^{2} = 1 \} \cup \{ (x,0) \colon -1\leq x \leq 1 \}.
\end{equation*}
Prove that $X$ is not homeomorphic to $S^{1}$.
My initial thought was to show that $X$ is not compact.  I was thinking that the collection of open balls with radius $r>0$, $\{ B(0,r) \}$, is an open cover with no finite subcover.  I am not very confident with my level of understanding so can someone tell me if I am on the right track or not.

You can also give an alternative proof using $\pi_1$. Since the theta-space is homotopic to the figure eight space i.e $\mathbb{S}^1 \vee \mathbb{S}^1$, it has fundamental group $\mathbb{Z} * \mathbb{Z}$. However, the fundament group of $\mathbb{S}^1$ is $\mathbb{Z}$. Since homeomorphic spaces have isomorphic fundamental groups, your result follows. 
