$ \vdash \alpha \to \alpha $
I'm trying to find a way solving this statement using minimal logic of Hilbert which have only two axiom's K & S and one only rule the modus pones (MP) :
what i've done so far :
If this is false correct me please , any other method to do it are welcome .
Yes, that's perfectly fine and well written up. You might be interested in noting that the implicational fragment of Hilbert style intuitionistic propositional logic is, up to the way of writing things, the same as typed combinatory logic: Propositions correspond to types, and (valid) proof trees correspond to (well-typed) combinator terms. In light of this correspondence, you have checked that the identity combinator $\textbf{I}$ can be expressed in terms of the $\textbf{S}$ and $\textbf{K}$ combinator through $\textbf{I} = \textbf{SKK}$ - this can be directly read off your cleanly presented proof tree. In the link to combinatory logic, it can be found in section Examples of Combinators