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Source. Rado (1949), the theorem in §4, part (i), statement on printed p. 341 and proof on p. 342 (canonical PDF).

Statement. Part (i) of the Theorem: If independent subsets $L,L'\subseteq M$ satisfy $|L|<|L'|$, then some $x\in L'\setminus L$ makes $L\cup\{x\}$ independent. The sets may be infinite or uncountable.

Proof. Suppose, to the contrary, that $L\cup\{x\}$ is dependent for every $x\in L'\setminus L$. For each such $x$, dependence by finite character gives a finite dependent $C_x\subseteq L\cup\{x\}$. The set $C_x$ must contain $x$, since every finite subset of $L$ is independent. Put $A_x=C_x\setminus\{x\}\subseteq L$. Then $A_x$ is independent, and finite-rank monotonicity and the dependence of $C_x$ give

$$r(A_x\cup\{x\})=r(A_x)=|A_x|. \tag{1}$$

For $x\in L'\cap L$, instead put $A_x=\{x\}$; equation (1) holds for this choice too. Choose all these finite supports simultaneously. This follows the source's treatment of common elements, whose ordered tuple then has a repeated entry.

For every finite $F\subseteq L'$, we claim the ordinary Hall inequality

$$\left|\bigcup_{x\in F}A_x\right|\ge |F|. \tag{2}$$

Indeed let $U=\bigcup_{x\in F}A_x$. This is a finite subset of $L$, so $r(U)=|U|$. For each $x\in F$, equation (1) and persistence of dependence give $r(U\cup\{x\})=r(U)$. Adjoining all finitely many elements of $F$ therefore leaves the rank unchanged. Since $F$ is an independent finite subset of $L'$, monotonicity yields

$$|F|=r(F)\le r(U\cup F)=r(U)=|U|,$$

which proves (2). This is the finite-support content of the source's argument with the finite exchange inequality (11).

Apply Lemma 2 to the family $(A_x)_{x\in L'}$ on the ambient set $L$, with the auxiliary rank $s(B)=|B|$ for finite $B\subseteq L$. Cardinality rank satisfies (R1)–(R3): adding a new element raises it by one, and an element that does not raise it is already present. Condition (2) is exactly the rank condition for this family. Lemma 2 gives pairwise distinct choices $\phi(x)\in A_x\subseteq L$ for every $x\in L'$. Thus $\phi:L'\to L$ is an injection, contrary to $|L|<|L'|$. The proposed augmenting element must exist. $\square$

The lemma is applied to cardinality rank, not to the original rank $r$. All unions evaluated by a rank before that application are finite. The full argument is relative to the finite representative theorem inside Lemma 2 and to the stated choice assumptions; it is distinct from finite augmentation alone.