Does unit sphere $S$ homeomorphic to $[0,1]$ Let $(\mathbb R^2, \|\cdot\|)$ be two dimensional normed space and $S=\{x\in\mathbb R^2 : \|x\|=1\}$ the unit sphere of $(\mathbb R^2, \|\cdot\|)$. Since $S$ bounded and closed  in $\mathbb R^2$ then it is compact. Does $S$ homeomorphic to the segment $[0,1]$ with usual topolgy? And if so then how to construct such an homeomorphism? 
 A: No, they are not. Suppose there exists a homeomorphism $f:\Bbb S^1\longrightarrow [0,1]$, let $p\in\Bbb S^1$ such that $f(p)\neq 0$ and $f(p)\neq 1$. Then $f:\Bbb S^1\setminus \{p\}\longrightarrow [0,1]\setminus\{f(p)\}$ should be still a homeomorphism. But the source is still connected whereas the target is not connected.
A: Another way of showing that $S^1$ is not homeomorphic to $[0,1]$ is by analyzing auto-homeomorphisms:

Definition. A topological space $X$ is called homogenous if for any $x,y\in X$ there exists a homeomorphism $f:X\to X$ such that $f(x)=y$.

The idea here is that in a way homogenous spaces have no special points.

Lemma. Let $X,Y$ be two homeomorphic spaces. Then $X$ is homogenous if and only if $Y$ is. In other words homogeneity is a topological invariant.

I leave the proof as an exercise.
With that you can easily see that $S^1$ is indeed homogenous: treat $S^1\subseteq\mathbb{C}$ and then for any $x,y\in S^1$ a homeomorphism that we are looking for can be for example $z\mapsto zx^{-1}y$ with complex multiplication on the right side.
Now $[0,1]$ is not homogenous. Indeed, if $f:[0,1]\to[0,1]$ is a homeomorphism then $f(0)=0$ or $f(0)=1$ as a consequence of the intermediate value theorem. So $0$ and $1$ are special points in $[0,1]$.
Note that this kind of argument can be extended to any dimension: an $n$-dimensional sphere is not homeomorphic to an $m$-dimensional cube because $n$-dimensional sphere is homogenous unlike $m$-dimensional cube (boundary points on a $m$-dimensional cube are special).
A: Another way to see that $S^1$ and $[0,1]$ are not homeomorphic using standard topology (i.e. not using homology/homotopy) is by using the Brouwer Fixed Point Theorem, which asserts that any continuous function $$f \colon [0,1] \to [0,1]$$ has at least one fixed point $x_0$.
But any non-zero rotation map on $S^1$ is continuous without any fixed points, for instance $$r \colon S^1 \to S^1, e^{i\theta} \mapsto e^{i \theta + \varphi},$$ with $\varphi \neq 0$.
