In the wikipedia article on Time Complexity it is written that:
The exponential time hypothesis (ETH) is that 3SAT, the satisfiability problem of Boolean formulas in conjunctive normal form with, at most, three literals per clause and with n variables, cannot be solved in time $2^{o(n)}$. More precisely, the hypothesis is that there is some absolute constant $c>0$ such that 3SAT cannot be decided in time $2^{cn}$ by any deterministic Turing machine.
We have $2^{o(n)} = \bigcap_{c > 0} O(2^{cn})$.
Proof: For the inclusion $2^{o(n)} \subseteq \bigcap_{c > 0} O(2^{cn})$. Let $f \in o(n)$, then we know that for each $c > 0$ we can find an $N_c$ such that for $n > N_c$ we have $f(n) < cn$, or $2^{f(n)} < 2^{cn}$ which gives this inclusion. For the inclusion $\bigcap_{c > 0} O(2^{cn}) \subseteq 2^{o(f)}$. Let be some function $g : \mathbb N \to \mathbb N \in \bigcap_{c > 0} O(2^{cn})$, i.e., for each $c > 0$ we have some $N_c$ such that for $n > N_c$ we have $g(n) < 2^{cn}$. Taking logarithms this gives $\log g(n) < cn$. So we have $\log g(n) \in o(n)$.
But ETH is about algorithms, so if we can find for each $c > 0$ some algorithm running in time $O(2^{cn})$, this does not imply that we have a single algorithm running in the intersection of those times, i.e., in $2^{o(n)}$ by the above. So there is still the possiblity that we cannot solve a problem in $2^{o(n)}$, but we can solve it by giving for each $c > 0$ some algorithm $A_c$ that solves it in $O(2^{cn})$. So claiming that something is not solvable in $2^{o(n)}$ is actually a stronger claim than saying we have a $c > 0$ such that the problem cannot be solved in $O(2^{cn})$.
But is there any example of such a problem, i.e. a problem such that for each $c > 0$ we have an algorithm (deterministic TM) running in time $O(2^{cn})$, but we cannot give an algorithm running in time $2^{o(n)}$?