$\mathbb{R}/{\sim}$ is compact for $x\sim y$ if $x=2^n y$ Let $\mathbb{R}$ with the standard topology and endow equivalence relation $x \sim y$ if $x=2^n y$ for $n \in \mathbb{Z}$.
I want to show $\mathbb{R}/ \sim $ is a compact space, and further for $\mathbb{R}^{+} = \{x \in\mathbb{R}\mid x>0\}$, $\mathbb{R}^+ / {\sim}$ is homeomoprhic to the unit circle $S^1$.

First I know $\mathbb{R}$ is not compact and $S^1$ is compact.  To show $\mathbb{R}/{\sim}$ is compact, I have to show its open cover has a finite subcover.  But I am not sure how this equivalence relation provides finite subcover.
 A: $\newcommand{\R}{\Bbb R}\newcommand{\RR}{\Bbb R/{\sim}}$ Let $\pi : \R \to \RR$ be the standard projection map.

For the first part, here's a sketch: Given any open set $U \subset \Bbb R$ containing $0$ and any $x \in \Bbb R$, we can choose $N \in \Bbb Z^+$ large enough such that $2^{-N}x \in U$.
Thus, $\pi(x) = \pi(2^{-N}x) \in \pi[U]$ for any $x \in \R$ and any open neighbourhood of $0$.
From this, conclude that the only open neighbourhood of $\pi(0)$ is $\RR$. In turn, every cover actually has a singleton subcover!

For the second part: Consider the map $$f : \R^+ \to S^1$$ defined as $$x \mapsto \exp(2 \pi \iota \log_2(x)).$$
Here $\iota$ denotes the imaginary unit and I consider $S^1$ as a subspace of $\Bbb C$ in the standard way.
It is easy to see that $f$ is onto. (Use the fact that $\log_2 : \R^+ \to \R$ is onto.)
For $x, y \in \R^+$, note the following
\begin{align}
f(x) = f(y) &\iff \exp(2 \pi \iota (\log_2(x) - \log_2(y)) = 1 \\
& \iff \log_2(x) - \log_2(y) \in \Bbb Z \\
& \iff x \sim y.
\end{align}
Thus, $f$ induces a continuous bijection $\tilde f : \R^+/{\sim} \to S^1$. (This is a fact about quotient spaces. For example, see Corollary 22.3. from Munkres.)
Now, to show that $\tilde f$ is a homeomorphism, it suffices to show that $f$ is a quotient map (again, the same Corollary from Munkres).
But this follows because $f$ can be seen to be a composition of the following maps:
$$\R^+ \xrightarrow{\log_2} \R \xrightarrow{x \mapsto \exp(2\pi\iota x)} S^1.$$
The first map is a homeomorphism (and hence, a quotient map) and the second is a quotient map. (This is a standard fact, you might want to prove it if you have not.)
Thus, being the composition of quotient maps, $f$ is also a quotient map and thus, $\tilde f$ is a homeomorphism.
