# Can the Lorentz transformations be derived this way?

Your derivation, up to before the point where you mention $C$, is the standard derivation. Once you have assumed the linearity of the transform, *i.e.* that $B$ has position ($\gamma t, \pm\gamma v t$), the standard derivation continues with the observation that the "length" (it's not positive semi-definite in Minkowski space) of a four-vector $x^\mu \leftrightarrow (t, x)$, defined (setting $c=1$) as
$$s^2 = t^2 - x^2,$$
is invariant under Lorentz transformation.

The proof that $\gamma^{-2} = 1 - v^2$ follows by calculating $s^2$ in the frame of $A$ and $B$ and equating these expressions. (The "other" sign, $\tilde\gamma^{-2} = 1+v^2$ does not preserve the four-length above.)

Invariance of this length is equivalent to the statement that $c=1$ for all observers. (An excellent, concise discussion is given throughout Ch. 2 of Rindler's *Essential Relativity*; see p.31ff, Sec. 2.6 of the *Revised Second Edition*, in particular.)

As commented by Frobenius, you have not used this fact anywhere in your derivation.

Your mention of 'faster than light inertial frames' is outside of the purview of Einstein's theoretical framework. In particular, you will violate the order of events for those at causal separation. Causality is a nice thing to maintain.