Defining the new proof system $N$ as this: We have 2 Axioms - $$A \rightarrow (A \lor B)$$ $$A \rightarrow (B \rightarrow A)$$
A new deduction rule: $$\bullet \frac{(A \rightarrow B)}{A \rightarrow (A \rightarrow B)}$$
We need to proof or disproof (formally):
$I)$ The statement $\alpha \rightarrow \alpha$ is a sentence in this system (basically showing that $\vdash_N \alpha \rightarrow \alpha$)
Now for the question itself is how do I prove this formally? I get the fact that I am always forced to have at least two logical connectives but how do I show that?
lets define $n(A)$ as the number of connectives in formula $A$.
By induction on the length of proof in $\mathsf{N}$ we show that if $\vdash_\mathsf{N} A$, then $n(A) \geq 2$.
If $A$ is an axiom, then it is either $A \rightarrow (A \lor B)$ or it is $A \rightarrow (A \rightarrow B)$. note that $n(A \rightarrow (A \lor B))=n(A \rightarrow (A \rightarrow B))=2n(A) + n(B) +2$ which is $\geq 2$. If $A$ is a theorem which is not an axiom, then it is derived by the rule. So $A$ is of form $B \rightarrow (B \rightarrow C)$, where $\vdash_\mathsf{N} B \rightarrow C$. So by induction hypothesis we have $n(B\rightarrow C) \geq 2$, hence $n(A)\geq 3$. qed.
Now because $n(p \rightarrow p)=1$ for atomic $p$, $\quad\nvdash_\mathsf{N} p \rightarrow p$.
Here I supposed that the system is given by axiom-schemes. You could argue similarly when we have substitution is present as a rule of inference and axiom are given as formulas.