Are $\{re^{i\theta}: 0Are $A := \{re^{i\theta}: 0<r\le 1, 0\le \theta < 2\pi \}$ and $B:=\mathbb{R} \times (0,1]$ homeomorphic?
My intuition tells me no. And yet, I can not find a single topological property that one has and the other doesn't.
 A: Given (from the comments) that using homotopy will not be helpful for the asker, here is a solution that avoids it. I assume as understood that $S^1$ and $\mathbb{R}$ are not homeomorphic. This is clear since the former is compact and the latter is not. I also assume as understood that $f$ restricts to a homeomorphism between a subspace $U$ of $A$ and the image $f(U) \subset B$ with the subspace topology. A reference for this fact is here.
Suppose, for sake of contradiction, there exists a homeomorphism $f : A \rightarrow B$. Denote by $U$ the set of points $x$ in $A$ such that there is an open neighborhood $V$ of $x$ for which $V\setminus \{x\}$ is homeomorphic to the punctured half-disk $D := \{(a,b) \in \mathbb{R}^2 : \lvert (a,b) \rvert < 1, (a,b) \not= (0,0) \mbox{ and } a \leq 0\}$. The homeomorphism $f$ must send $U$ to the analogously defined subset in $B$ (otherwise it does not restrict to a homeomorphism on $V\setminus \{x\}$), and vice versa for $f^{-1}$. Hence $f$ should restrict to a homeomorphism $f\rvert_U : U\rightarrow f(U) \subset B$ (with $U$ and $f(U)$ endowed with the subspace topologies) between these specific sets. However, $U \cong \mathbb{S}^1$ whereas $f(U)\cong \mathbb{R}$. This is a contradiction.
Hence $A \not\cong B$.
