Let $X_1$ and $X_2$ be the spaces \begin{align*} X_1&=\{(x,y)\in \mathbb{R}^2 : (x,y)\neq(0,0)\}, \\ X_2&=\{(x,y)\in \mathbb{R}^2 : (x,y)\notin [0,1]\times\{0\}\}. \end{align*} Are these spaces homeomorphic? If these spaces are homeomorphic, what is the homeomorphism map? If these spaces are not homeomorphic, why?
I suspect these spaces are homeomorphic, but I can't construct homeomorphism.
On
On $A=I\times([-1,1]\setminus\{0\})$ let $f:A\to\Bbb R^2$ be the map $$f(x,y)=(x|y|,\,y)$$ on $B=[1,\infty)\times[-1,1]\setminus\{(1,0)\}$ let $f:B\to\Bbb R^2$ be the map $$f(x,y)=(x-(1-|y|),\,y)$$ and on $C=\Bbb R^2\setminus\left((0,\infty)\times(-1,1)\cup\{(0,0)\}\right)$ let $f:C\to\Bbb R^2$ be defined by $$f(x,y)=(x,y)$$ Can you show that $f:X_2\to X_1$ is continuous and find its inverse?
You are right with your conjecture; but there is no such thing as "the" homeomorphism between these two spaces. In fact you have to put one together from your geometrical and analytical toolkit.
Here is an idea. I'm going to replace the segment $[0,1]$ in the definition of $X_2$ by $[-1,1]$, so that the formulas get simpler. We can cover $X_2$ by a family of confocal ellipses with foci $(\pm1,0)$. A typical such ellipse has parametric representation $$\gamma_b:\quad t\mapsto\bigl(\sqrt{b^2+1}\cos t,\>b\sin t\bigr)\qquad(0\leq t\leq 2\pi)$$ with $b>0$. Now set up your homeomorphism in such a way that $\gamma_b$ is mapped onto a circle of radius $b$.
As @drhab has uttered doubts I shall proceed with the details: Let $f:\>X_1\to X_2$ be such that $$(b\cos t,\>b\sin t)\mapsto\bigl(\sqrt{b^2+1}\cos t,\>b\sin t\bigr)\ .$$ Given $(x,y)=(b\cos t,\>b\sin t)$ we have $$ \cos t={x\over b},\quad \sin t={y\over b},\quad b=\sqrt{x^2+y^2}\ .$$ It follows that $$\sqrt{b^2+1}\cos t={x\sqrt{x^2+y^2+1}\over \sqrt{x^2+y^2}},\quad b\sin t= y\ .$$ This means that our homeomorphism appears in cartesian coordinates as $$f:\quad (x,y)\mapsto\left(x\sqrt{1+{1\over x^2+y^2}}, \ y\right)\ .$$