Propositional Logic, proving that the sequent of every valid argument can be proved I am studying a from "A First Course in Logic" by Mark V. Lawson. I am currently on the first chapter which covers propositional logic. The author has just introduced Sequential Calculus and states the following theorem regarding completeness (Page 100):

Theorem 1.12.4 (Completeness)- Let $A_{1}, \dots, A_{m} \vdash B_{1}, \dots, B_{n}$ be a valid argument. Then the sequent $A_{1}, \dots, A_{m} \Rightarrow B_{1}, \dots, B_{n}$ can be proved.

The argument used by the author is as follows:

Let $\mathbf{U} \Rightarrow \mathbf{V}$ be a sequent. We prove that if there is a truth assignment $\tau$ falsifying the sequent $\mathbf{U} \Rightarrow \mathbf{V}$ then in any deduction tree for $\mathbf{U} \Rightarrow \mathbf{V}$ there is a leaf $\mathbf{X} \Rightarrow \mathbf{Y}$ such that $\tau$ falsifies $\mathbf{X} \Rightarrow \mathbf{Y}$, and conversely.

The rest of the proof is dedicated to showing this fact is true. However, I do not understand why this is sufficient in proving the theorem. The claim made seems to be a statement about sequents that aren't valid, rather than ones that are? Since a valid sequent has no falsifying truth assignments, this means there is no leaf $\mathbf{X} \Rightarrow \mathbf{Y}$ which can be falsified. However, I do not see how this guarantees that the leaves on any completed deduction tree will be axioms (in turn implying that the complete deduction tree is a proof tree).
Note that bold letters refer to sets of well founded formulae, e.g. $\mathbf{U} = \{A_{1}, A_{2}, \dots, A_{m}\}$, and the author defines an axiom as any sequent of the form $\mathbf{U}, X \Rightarrow \mathbf{V}, X$, where $X$ is a well formed formulae appearing on both sides of the sequent.
This my attempt at resolving my issues so far:
Assume $\mathbf{U} \Rightarrow \mathbf{V}$ is a valid sequent. Assume a leaf of the deduction tree $\mathbf{X} \Rightarrow \mathbf{Y}$ is not an axiom. Thus $\mathbf{X}\cap\mathbf{Y} = \emptyset$. This in turn implies that there must be a well-formed formula in either $\mathbf{X}$ or $\mathbf{Y}$ involving a negation, conjunction or disjunction.  If this was not the case, a falsifiable example would clearly exist. As a result, there is a rule that can be applied to $\mathbf{X} \Rightarrow \mathbf{Y}$ to increase the size of the deduction tree. This process can be repeated over and over. If the number of well founded formulae in both $\mathbf{U}$ and $\mathbf{V}$ are finite then this process will eventually terminate and the leaves of the deduction tree must be axioms.
This approach seems to work if as long as the number of well found formulae included in the sets $\mathbf{U}$ and $\mathbf{V}$ are finite (which I would assume is a necessary assumption in propositional logic?).
Is my argument and line of thinking correct, or am I missing something?
 A: In sequent calculus, every sequent on a line of derivation is a conditional tautology/theorem, the antecedent (possibly empty), $A_{1},\dots, A_{m}$ on the left-hand side of $\Rightarrow$ and the succedent (possibly empty), $B_{1},\dots, B_{n}$ on the right-hand side of $\Rightarrow$.
In order to prove completeness (the converse of soundness), we have to show that  any sequent $\mathbf{U}\Rightarrow\mathbf{V}$ can be decomposed until we reach axioms (in the book's, hence, Smullyan's version). All axioms are of the form $\mathbf{U}, X\Rightarrow\mathbf{V}, X$. We shall deal with a non-axiom sequent, otherwise, the task has been done. So, $\mathbf{U}$ and $\mathbf{V}$ have no common wff. Therefore, we can assign truth-values to them independently from one another. Also we establish that (as Question 3 of Exercises 1.12; I have renamed the sets for the sake of uniformity)

There is a truth assignment $\tau$ falsifying the sequent
$\mathbf{W}\Rightarrow \mathbf{Z}$
if and only if
in any deduction tree for $\mathbf{W}\Rightarrow\mathbf{Z}$ there is a
leaf $\mathbf{U}\Rightarrow\mathbf{V}$ such that $\tau$ falsifies
$\mathbf{U}\Rightarrow\mathbf{V}$

We suppose $\mathbf{W}\Rightarrow\mathbf{Z}$ is a sequent corresponding to a valid argument. By applying the decomposition rules, we reach a leaf a non-axiom sequent $\mathbf{U}\Rightarrow\mathbf{V}$ occupies. Since they are independent, we arrange $\tau$ such that it assigns truth to the atomic formulas in $\mathbf{U}$ and falsity in $\mathbf{V}$ —see the exercise how this results in that the antecedent $\mathbf{U}$ true while the succedent $\mathbf{V}$ is false. Thus, by the initial part, invalidity of the sequent $\mathbf{U}\Rightarrow\mathbf{V}$ is "transmitted" to the sequent $\mathbf{W}\Rightarrow\mathbf{Z}$, which contradicts our assumption.
Therefore, $\mathbf{U}\Rightarrow\mathbf{V}$ has to be an axiom. Since we can translate its deduction tree into a proof reading from the downside up, we have shown that any sequent $\mathbf{W}\Rightarrow\mathbf{Z}$ for a valid argument is provable starting off with axioms.
