I want to prove that each $\Gamma$ with $\Gamma \vdash \varphi$ contains a finite $\Delta$, such that also $\Delta \vdash \varphi$.
This is one of my homework problems and as a hint it says that we can use induction on the derivation tree, but I don't see any need to use induction. If $\Gamma \vdash \varphi$ then by definition there is a derivation with conclusion $\varphi$ and with all (uncanceled) hypotheses in $\Gamma$. Number of leafs of this tree is finite, and by combining all uncanceled leafs, we can make the set $\Delta$ which is a subset of $\Gamma$ and the same derivation shows that $\Delta \vdash \varphi$. What am I missing?
You are right, no induction needed. Every derivation in classical logic is of finite length, which means you can only have used a finite number of premises.
Maybe the induction is meant to be used that every derivation is of finite length? That is, since you start with having 0 derived statements, and since you only add one more derived statement at a time, your derivation is always of finite length.
Make sure to point out that each step in the derivation refers to only a finite number of earlier statements (one can imagine logics where one can derive in one step a statement from an infinite number of statements, but that is not classical logic or, more to the point: that is presumably not what any of the rules in the system that you have to use allow you to do). Thus: a finite number of steps referring to a finite number of earlier statements can only refer to a finite number of premises, and that is your finite subset $\Delta$.