Big O proving for increasing functions
The claim is false. Let $f(n) = n^{1.75}$ and $g(n) = 2^{\lfloor \log_2 n \rfloor} n$, and note that both are increasing. When $n = 2^t$, we have $g(n) = n^2$, while when $n = 2^t-1$, we have $g(n) \approx n^{1.5}$. This shows that
$$
\liminf_{n\to\infty} \frac{f(n)}{g(n)} = 0, \quad
\limsup_{n\to\infty} \frac{f(n)}{g(n)} = \infty.
$$
In particular, it is not the case that $f = O(g)$, nor is it the case that $g = O(f)$.