Let the quotient space be $[0,1]\times[0,1]/\sim_{1}$, where $\sim_{1}$ is defined as $(0,t)\sim_{1}(1,1-t)$. We say the given quotient space is the Mobius band. Now we need to prove this quotient space is homeomorphic to $S^1\times[0,1]/\sim_{2}$, where $(x,1)\sim_{2}(-x,1)$.
I think to prove the two quotient spaces are homeomorphic we need to find a homeomorphism $f:[0,1]\times[0,1]\to S^1\times[0,1]$ such that $x\sim_1 x'$ iff $f(x)\sim_2 f(x')$. The problem is that I can't find that. Thanks for assistance in advance.
It is impossible to find a homeomorphism $f:[0,1]\times[0,1] \rightarrow \mathbb{S}^1\times [0,1]$, since the former is contractible, while the latter is homotopy equivalent to $\mathbb{S}^1$ and thus not contractible.
But the idea is right. I am not sure, whether the equivalence relation $\sim_2$ is actually correct in the sense that it is not clear to me which map you are supposed to construct. In the following I thus prove a slightly different statement and you can adapt it to your exercise, in case $\sim_2$ makes sense.
First prove that the space $[0,1]\times[0,1]/\sim_1$ is Hausdorff. Then check that the map $$\begin{array}{rcl} f:\mathbb{S}^1 \times [0,1]&\rightarrow& [0,1]\times[0,1]/\sim_1\\ (e^{2\pi it},s) & \mapsto & \left\{\begin{array}{ll}(2t,s)&t\in[0,\frac{1}{2}]\\(2t-1,1-s)&t\in [\frac{1}{2},1]\end{array}\right. \end{array}$$ gives a welldefined, surjective, continous and by the compact-Hausdorff-trick open map. This map identifies $(e^{2\pi it},s) \sim_3 (e^{2\pi i(t+\frac{1}{2})},1-s)$ or if you wish $(x,s) \sim_3 (-x,1-s)$. Thus we get an induced homeomorphism $$\widetilde{f}:\mathbb{S}^1\times[0,1]/\sim_3 \;\overset{\cong}{\longrightarrow}\; [0,1]\times[0,1]/\sim_1$$ (it is injective and continuous by the universal property of the quotient and surjective and open since $f$ is).