Possibility to prove that a function in big-theta is in big-O and little-o
Yes, it looks like that you missed something obvious.

A stronger claim: If $f(n) = \Theta (g(n))$ and $g(n) = o(h(n))$ then $f(n) = o(h(n))$.



Sketch of a proof: $f$ is bounded both above and below (up to a constant factor) by $g$ asymptotically, which is informally understood as $f$ and $g$ are the same asymptotically. $g$ is dominated by $h$ asymptotically. Hence $f$ is dominated by $h$ asymptotically. 

(Sure, the above sketch is no more than rephrasing the claim in English. Well, that is how you (or at least I) would like to understand and reason about the situation)



Here is a detailed proof.

$f(n) = \Theta (g(n))$ implies that $c_1g(n) \le f(n) \le c_2 g(n)$ for some $c_1\gt0,\,c_2\gt0,\,k_1$ and any $n \gt k_1$.

$g(n) = o(h(n))$ implies that $|g(n)| &lt; c_3 h(n) $ for some $c_3\gt0,\,k_2$ and any $n \gt k_2$.

Let $k_3=\max(k_1,k_2),\, c_4=\max(c_1,c_2)c_3\gt0$. Then for any $n\gt k_3$,
$$\begin{align}
|f(n)|&amp;\le \max(|c_1g(n)|, |c_2g(n)|)\\
      &amp;\lt\max(c_1c_3h(n),c_2c_3h(n))\\
      &amp;=\max(c_1,c_2)c_3h(n)\\
      &amp;=c_4h(n)
\end{align}$$
Note the first inequality above is guaranteed by the fact that $f(n)$ is bounded between $c_1g(n)$ and $c_2g(n)$ since $n&gt;k_1$.



Finally, note that $f(n)=o(h(n))$ implies $f(n)=O(h(n))$.



P.S. There is some typo in your question. For example, "$c_1, c_2\in\mathbb{N}$" should be "some $c_1,\,c_2\in\Bbb{R}^+$" or "some $c_1&gt;0,\,c_2&gt;0$". If you let $f(n)=n, g(n)=2n$, then there is no natural number $c_1$ such that $c_1g(n)\le f(n)$ for even one positive $n$.