Proof that $S^1$ isn't homeomorphic to $[0, 1]$

Question

Prove that $S^1$ isn't homeomorphic to $[0, 1]$

So when I tried to prove this, I tried looking at topological properties that $S^1$ had but $[0, 1]$ didn't because trying to prove it by contradiction seemed a lot harder, and I'm not sure how I could prove it directly.

The only topological properties that $S^1$ has that I can think of that $[0,1]$ doesn't have is the fact that $S^1$ is a topological manifold without boundary and $[0, 1]$ is a topological manifold with boundary, that being the set $\{0, 1\}$ so if we assume they're homeomorphic we'd arrive at a contradiction since a topological manifold with boundary can't be homeomorphic to a topological manifold without boundary.

But that seems a bit overpowered to use topological manifold theory to prove this somewhat elementary problem. Does anyone know a more elementary proof?

Answer 1

Take any $z\in S^1$. Then $S^1\setminus\{z\}$ is connected, while this property does not hold for $[0,1]$ unless we remove the boundary points.

Answer 2

No neighborhood of $0$ in $[0,1]$ is homeorphic to $\mathbb R$. However, every point of $S^1$ has some neighborhood homeorphic to $\mathbb R$.