Hardness of a subcase of Set Cover How hard is the Set Cover problem if the number of elements is bounded
by some function (e.g, $\log n$) where $n$ is the size of the problem instance. Formally, 

Let $\mathcal{U}=\{e_1, \cdots, e_m\}$ and $\mathcal{F} = \{S_1, \cdots, S_n\}$
where $S_i \subseteq \mathcal{U}$ and $m = O(\log n)$. How hard is it to decide the following problem 

\begin{align*}
         \text{SET-COVER'}=\{&lt;\mathcal{U}, \mathcal{F}, k&gt;:
         &amp;\text{ there exists at most $k$ subsets }\\
         &amp;\text{ $S_{i_1}, \cdots, S_{i_k}\in \mathcal{F}$ that cover $\mathcal{U}$}
         \}.
\end{align*}

What if $m=O(\sqrt{n})$? 

Any result based on well known conjectures (e.g., Unique Games, ETH) is good. 

Edit 1: A motivation for this problem is finding out when the problem gets hard as $m$ increases. Clearly, the problem is in P if $m=O(1)$ and NP-hard if $m=O(n)$. What is the threshold for the NP-hardness of the problem?

Edit 2 : There exists a trivial algorithm to decide it in time $O(n^{m})$ (which enumerates all subsets of size $m$ of $\mathcal{F}$). Therefore, the problem is not NP-hard if $m=O(\log{n})$ since ETH implies that there is no algorithm in time $O(2^{n^{o(1)}})$ for any NP-hard problem (where $n$ is the size of the NP-hard problem).
When $m = O(\log n)$, you can use dynamic programming to find the optimum in polynomial time. The table contains Boolean-valued cells $T_{\ell,X}$ for each $\ell \in \{0,\ldots,k\}$ and $X \subseteq \mathcal{U}$, indicating whether there are $\ell$ sets which cover the elements in $X$.

When $m = O(\sqrt{n})$, say $m \leq C\sqrt{n}$, the problem remains NP-hard. Given an instance of SET-COVER, add $m$ new elements $x_1,\ldots,x_m$ and $(2C^{-1}m)^2$ new sets, consisting of non-empty subsets of the new elements, including $\{x_1,\ldots,x_m\}$ (when $m$ is large enough, $(2C^{-1}m)^2 &lt; 2^m$). Also increase $k$ by one. The new $m,n$ are $m' = 2m$ and $n' = n + (2C^{-1}m)^2 \geq (C^{-1}m')^2$.