Show that the circle $\{(x, y) ∈ R^2 : x^2 +y^2 = 1\}$ is not homeomorphic to the closed unit interval $[0,1] ⊆ \mathbb R$. I am required to show that the circle $\{(x, y) ∈ \mathbb{R}^2 : x^2 +y^2 = 1\}$ (with the subspace topology induced from the usual one in $\mathbb{R}^2$ ) is not homeomorphic to the closed unit interval $[0,1] ⊆ \mathbb{R}$.
So far I have: 
Suppose for a contradiction there exists a homeomorphism between $[0,1]$ and $S^1$, then there is a homeomorphism between $[0,1]\setminus \{\frac{1}{2}\}$ and $S^1 \setminus \{f(\frac{1}{2})\} $. I am trying to justify now that the domain is now disconnected but the codomain is still connected, and this cannot hold is two sets are homeomorphic to each other. How would one correctly argue that the domain is disconnected? Are my arguments sound until this point?
Thanks in advance.
 A: This is a basic property of intervals: if $\;I\;$ is any interval and $\;a\;$ is any interior point, then $\;I\setminus\{a\}\;$ is no longer an interval (and thus disconnected).
Or also Define
$$f:[0,1]\setminus\left\{\frac12\right\}\to\left\{0,\,1\right\}\;,\;\;f(x)=\begin{cases}0,\,&0\le x<\frac12\\{}\\1,\,&\frac12<x\le1\end{cases}$$
Then $\;f\;$ is a continuous surjection and thus the domain must be disconnected (the set $\;\{0,1\}\;$ with the inherited topology from $\;\Bbb R\;$ , i.e. it is a discrete space with two points)
A: Your argumentation is sound; The only thing missing are indeed proofs of the domain being disconnected and the codomain being connected. I will provide them below.

After removing the point $\frac{1}{2}$ from $I = [0,1]$ you are left with $I' = [0, 1] \setminus \{ \frac{1}{2} \}$. 
$I'$ is disconnected: It can be written as the disjoint union of $[0, \frac{1}{2})$ and $( \frac{1}{2}, 1]$, which are both open in $I'$ since they are open in $I$.
However, $S' := S^1 \setminus \{ f(\frac{1}{2}) \}$ is path-connected and hence connected: Any two points $a,b \in S'$ can be written in polar coordinates
$$
a = (\cos \alpha, \sin\alpha) ~~,~~
b = (\cos \beta, \sin\beta) ~~,~~
f(\tfrac{1}{2}) = (\cos \gamma, \sin\gamma) 
$$
with $\alpha, \beta, \gamma \in [0, 2\pi)$.
Based on this, we can define a continuous path in $S^1$ connecting $a$ and $b$. If $\alpha \leq \gamma \leq \beta$, we take the path
\begin{align}
p : [0,1] \rightarrow S^1 ~&;~ \theta \mapsto  
\big(
~\cos(\beta + \theta(2\pi - \beta + \alpha) ), ~\sin(\beta + \theta(2\pi - \beta + \alpha ) 
~\big)
\end{align}
and otherwise
\begin{align}
p : [0,1] \rightarrow S^1 ~&;~ \theta \mapsto  
\big(
~\cos(\alpha + \theta(\beta - \alpha) ), ~\sin(\alpha + \theta(\beta - \alpha) 
~\big)
\end{align}
In either case, $p$ is continuous and does not contain $f(\frac{1}{2})$, so it is also a continuous path between $a$ and $b$ if we restrict it to $S'$.
