Proving that every derivation-tree has at most one leftmost-derivation in a context free grammar
Here is a paraphrase of your proof. Suppose
$$
S \Rightarrow \alpha_1 \Rightarrow \cdots \Rightarrow \alpha_n \Rightarrow w \\
T \Rightarrow \beta_1 \Rightarrow \cdots \Rightarrow \beta_m \Rightarrow w
$$
are two leftmost derivations corresponding to the same derivation tree. Then the first step must be the same. This is the point where you're stuck.

What you want to do now is to say that $\alpha_1 \Rightarrow \cdots \Rightarrow w$ and $\beta_1 \Rightarrow \cdots \Rightarrow w$ are also leftmost derivations corresponding to the same derivation tree (obtained from the original one by taking into account the first step), and so by induction these derivations are the same.

There is a small problem – $\alpha_1$ is now not a single symbol! You therefore have to strengthen your induction hypothesis and prove something more general which involves derivation trees starting from sentential forms (words which can involve both terminals and non-terminals).



Alternatively, if you prefer using frontiers of derivation trees, you should strengthen your induction hypothesis in a different (but equivalent) way. Now you are still given the original derivation tree, but some of the productions are marked as already having been performed. Your induction hypothesis now states that all leftmost derivations from this state are the same.



It's also possible to avoid strengthening the induction hypothesis, at the cost of a more complicated argument during the inductive step. Suppose that the first step is $S \to x_1 A_1 x_2 A_2 \dots x_n A_n x_{n+1}$, where $A_1,\ldots,A_n$ are non-terminals and $x_1,\ldots,x_{n+1}$ are words. Any leftmost derivation must first completely derive $A_1$, then completely derive $A_2$, and so on, until $A_n$ is completely derived. Induction shows that there is only one way to achieve any of these $n$ steps.



More generally, I suppose you focus first on why the result is true; only then focus on how to write the argument formally.