Prove $S^1$ and $\{(x,y) : 1 < x^2+y^2 < 4 \}$ are not homeomorphic using topological invariance of path components.
For this problem I refer to $S^1=\{(x,y) \in \mathbb{R}^2 : x^2+y^2=1\}$, i.e. the unit circle, and $\pi(X)$ to be the set of all path components of a set $X$.
Proof: Let $S^1$ be defined as above, and $T \subset \mathbb{R}^2$ to be the annulus defined as $T=\{(x,y):1<x^2+y^2<4\}$. Assume for the sake of contradiction that $f:S \longrightarrow T$ is a homeomorphism. Let $A=S^1 \setminus\{(1,0),(-1,0)\}$. Since $A \subset S^1$ we know that $A$ and $f(A)$ must be homeomorphic, and since $f$ is a homeomorphism it is also a bijection. Therefore we will have, $$f(A)=f(S^1 \setminus \{(1,0),(-1,0)\} = T \setminus \{f(1,0),f(-1,0)\}$$ Since $T$ is an annulus, it is not possible to remove any two distinct points and obtain an object which is not path connected. In other words, $|\pi(f(A))|=1$. However, notice that $A$ is homeomorphic to $(0,1)\cup(1,2) \subset \mathbb{R}$ which clearly has 2 path components. Hence, $|\pi(A)|=2$, a contradiction of the topological invariance of path components under homeomorphism. Conclude that $S^1$ and $T$ are indeed not homeomorphic. $\Box$
Does this proof contain any errors? Is there a more formal way to show that $f(A)$ must necessarily be path connected?