Can we show that every subgroup of a free group is free using Kurosh's theorem? I have seen classic proofs about " every subgroup of a free group is free".
My question now is 

Can we use Kurosh's theorem to prove the statement?

Kurosh's theorem may be found in the book: "Introduction to group theory," by Oleg Bogopolski, page 92. Thanks for any feedback!
 A: The answer to your question is "yes", as @QiaochuYuan suggests in the comments. Let me give a detailed answer. But, after that, I will express some reservations about this as a method of proof.
Consider a free group $F$. As QiaochuYuan suggests, write $F$ as a free product $F = \bigstar_{i\in I} \, Z_i$ where each group $Z_i$ is infinite cyclic. Given a subgroup $A < F$, Kurosh's Theorem let's you conclude that the group $A$ is a free product of the form $A = (\bigstar_{j\in J} \, A_j) * B$ with the following properties:


*

*The free factors $A_j$ ($j \in J$) are all of the nontrivial subgroups of $A$ that can be written in the form $A \cap g Z_i g^{-1}$ for $i \in I$ and $g \in F$. It follows that each $A_i$ is a nontrivial subgroup of an infinite cyclic group, and so is itself infinite cyclic.

*The subgroup $B$ is a free group.


Thus $A$ is a free product of infinite cyclic groups and one free group, from which one can conclude that $A$ is a free group.
So, what are my reservations? Roughly speaking, in the topological approach to studying free groups, one deduces that $B$ is a free group in pretty much the same way as one proves that a subgroup of a free group is free. Namely, in both cases, one arrives at the conclusion that the group in question is free by constructing a free action of that group on a simplicial tree; then one proves that any group that acts freely on a simplicial tree is a free group. For $B$ itself, the tree that one uses is the Bass-Serre tree of the free product decomposition $F = \bigstar_{i \in I} \, Z_i$. For a subgroup of a free group, the tree that one uses is the Cayley tree of the free group with respect to a free basis.
In a nutshell: the proof of the Kurosh subgroup theorem can be regarded as an extension of the proof that subgroups of free groups are free, an extension which requires the additional machinery of Bass-Serre theory to carry out in full. So it is technologically less expensive to just prove that subgroups of free groups are free than to deduce that from the Kurosh subgroup theorem.
