An attempt at proving that $A=(0,1)$ is not compact on the real line with the usual topology. I am supposed to show that the open interval $(0,1)$ on the real line $\mathbb{R}$ (with the usual topology) is not compact. I know that one of the examples of an open cover without a finite subcover, in this case, would be $A=\left\{\left(\dfrac{1}{n},1\right)\bigg\rvert n\in\mathbb{N}\right\}$, thus proving that $A$ is not compact.
However, I am trying to prove the same using a different example of an open subcover, $$\mathscr{F}=\left\{\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)\bigg\rvert n\in\mathbb{N}\right\}.$$ If it can be shown that the family of sets $\mathscr{F}$ forms an open cover for $(0,1)$, i.e., $\bigcup\limits_{n\in\mathbb{N}}G_n$, where $G_n=\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)$, it is straightforward to check that for any $x=\dfrac{1}{2^n}$, where $n\in\mathbb{N}$, it only lies in one of these $G_n$'s. So, removing any one of these open sets from our open cover, in an attempt to find a finite subcover, the family of sets won't form a finite subcover of $(0,1)$. Thus, $(0,1)$ is not compact.
Where I am getting stuck is in actually formally proving that any $x\in(0,1)$ lies in one of these $G_n$'s. My intuition here is that by the use of Archimedean principle, for any $x\in(0,1)$, there exists some $k\in\mathbb{N}$ such that $x>\dfrac{1}{k}$. So, let us choose $k$, for a given $x$, such that $\dfrac{1}{k}<x<\dfrac{1}{k-1}$. Again, since this $k>1$, there exists some $n_0\in\mathbb{N}$ such that $2^{n_0-1}<k<2^{n_0}\implies \dfrac{1}{2^{n_0}}<\dfrac{1}{k}<\dfrac{1}{2^{n_0-1}}$. My claim is that $x\in \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, and approach was to try and prove that $\left(\dfrac{1}{k},\dfrac{1}{k-1}\right)\subseteq \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, but I am unable to proceed ahead from this point.
Any help would be highly appreciated, in either proving or disproving the claim that I make in the last sentence.

Let $0<x<1$. Consider the interval $(\frac {\ln (\frac 1  x)} {\ln 2},\frac {\ln (\frac 3  x)} {\ln 2})$. The length of this interval is easily seen to be $\frac {\ln 3} {\ln 2}$. Since any interval of length greater than $1$ contains an integer there exist a positive integer $n$ in this interval. You can now verify that $x \in (\frac 1 {2^{n}},\frac 3 {2^{n}})$. Note that since $0 <x<1$ and $x \in (\frac 1 {2^{n}},\frac 3 {2^{n}})$ the integer $n$ is necessarily a positive integer.
