Say, an equation is given below \begin{equation} 2\pi(x) - \pi(2x)=\omega(x) \end{equation} where $x$ is a positive integer, $\pi(x)$ is the prime-counting function, and $\omega(x)$ is the number of distinct prime factors of $x$. I would like to know if this equation holds for finitely many values of $x$ or infinitely many.
If one uses asymptotic formula (from number theory) it can be written that \begin{equation} \omega(x) \sim \ln x / \ln \ln(x) \end{equation} (from http://www.math.uiuc.edu/~hildebr/ant/main3.pdf) and \begin{equation} \pi(x) \sim x/\ln(x), \end{equation} so the above equation can be rewritten (after simplification) as \begin{equation} 2x\ln(2)=\ln(x)\ln\ln(x)\ln(2x) +f(x) \end{equation} where $f(x)$ is an error term.
If I plot left side (which is linear, supposed to be a straight line) and right side(a curve),assuming $ f(x) $ is 100% accurate, then the equation implies(?!) that the right will intersect left side. Since line can intersect a curve ($ln $ function) finite times, so there are finite intersection point. So, can I say that there are only finitely many values of $x$?
graph example-

here f(x) is assumed as x + sinx
without the f(x)-

Your proof is wrong, as Erick Wong has explained. Here is a proof of your assertion:
The function $2\pi(x)-\pi(2x)$ is asymptotic to $2x\log 2/\log^2x$ by the Prime Number Theorem (as proved with a stronger error term than just $\sim$). $\omega(n)$ has maximal order $\log n/\log\log n$ on the primorials, and $$ \lim_{n\to+\infty}\frac{2n\log 2/\log^2n}{\log n/\log\log n}=+\infty. $$
As a result there can be only finitely many solutions $2\pi(n)-\pi(2n)=\omega(n).$