Is following observation on Ladner's theorem correct?
I'm annoyed at both ["any" being used as a quantifier] and [the placement of quantifiers making things even more ambiguous], so I'll start with the two interpretations of your initial block-quote.



  Suppose there is a function $\hspace{.04 in}f$ in $\omega(1)$ such that ​ NP $\subseteq$ DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}f(n)}\hspace{-0.03 in}\right]$ .
  
  Does it follow that ​ P = NP ?


It's not known to, since ​ ​ ​ ($n\mapsto 2^{\hspace{.02 in}n}\hspace{-0.03 in}$) ​ $\in$ ​ $\omega(1)$ ​ ​ ​ and ​ NP $\subseteq$ EXP $\subseteq$ DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}(2^{\hspace{.02 in}n})}\hspace{-0.04 in}\right]$ ​ are both known.



  Suppose that for all functions $\hspace{.04 in}f$ in $\omega(1)$, ​ NP $\subseteq$ DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}f(n)}\hspace{-0.03 in}\right]$ .
  
  Does it follow that ​ P = NP ?


Yes, since as described in this answer, there exists

a function $\hspace{.04 in}f$ in $\omega(1)$ such that ​ DTIME$\left[\hspace{-0.02 in}n^{\hspace{.04 in}f(n)}\hspace{-0.03 in}\right]$ = P .

The truth of your "That is ... fixed $c\geq 1$." sentence has nothing to do with Ladner's Theorem

- The empty language and full language, for example, are such NP problems.

For the sentence after that, the g given by ​ g(n) = c ​ is such a g, and functions f cannot simultaneously be in $\omega(1)$ and satisfy ​ f(n) &lt;&lt; c , ​ so where's the violation of Ladner's theorem?

By the function g I just mentioned, the part of your

"And so if ..." sentence before that sentence's "and" is true.

For the rest of that sentence, see this answer's two block-quotes and its responses to those.



This answer's response to its "there is" block-quote would still apply.

I believe this answer's response to its "for all" block-quote

would also still apply, by a similar argument.