The regulator is well defined

Rank zero: if $r=0$ then by [F1] the matrix $A$ has no columns and each $A_k$ is the empty matrix with determinant $1$, while by [F2] the empty tuple is the unique system of fundamental units; hence $R_K=1$ is independent of the deleted row and of the fundamental system, and $R_K>0$.

Assume $r\ge 1$ from here on, so $A$ is an $(r+1)\times r$ real matrix with $r_1+r_2=r+1\ge 2$ rows. Its columns are $\mathbb{R}$-linearly independent, that is $A$ has rank $r$: by [F4] the group $\lambda ({\mathcal{O}}_K^{\times })$ is a discrete subgroup of the finite-dimensional real vector space $H$, so by [F5] it equals $\mathbb{Z}{v}_1\oplus \cdots \oplus \mathbb{Z}{v}_s$ with $v_1,\dots ,v_s$ $\mathbb{R}$-linearly independent and ${\mathrm{span}}_{\mathbb{R}}\lambda ({\mathcal{O}}_K^{\times })=\mathbb{R}{v}_1\oplus \cdots \oplus \mathbb{R}{v}_s$ of dimension $s$; by [F2] the elements $\lambda ({\epsilon }_1),\dots ,\lambda ({\epsilon }_r)$ form a $\mathbb{Z}$-basis of that same group, so $s=r$ (equal free rank) and their $\mathbb{R}$-span is all of ${\mathrm{span}}_{\mathbb{R}}\lambda ({\mathcal{O}}_K^{\times })$, of dimension $r$; a spanning set of $r$ vectors in a space of dimension $r$ is a basis, so the columns of $A$ are $\mathbb{R}$-linearly independent and the rank of $A$ is $r$ by [F7].

Every column of $A$ lies in $H$, hence has coordinate sum zero.

Independence of the fundamental system: let $({\epsilon }_1^{\mathrm{\prime }},\dots ,{\epsilon }_r^{\mathrm{\prime }})$ be a second system of fundamental units and let $A^{\mathrm{\prime }}$ be the matrix with columns $\lambda ({\epsilon }_1^{\mathrm{\prime }}),\dots ,\lambda ({\epsilon }_r^{\mathrm{\prime }})$. By [F2] both logarithm lists are $\mathbb{Z}$-bases of the same group. For each $i$, write $\lambda ({\epsilon }_i^{\mathrm{\prime }})={\sum }_j{b}_{ji}\lambda ({\epsilon }_j)$ using unique integers $b_{ji}$, and let $B=(b_{ji})$, so the $i$-th column of $B$ records the coordinates of $\lambda ({\epsilon }_i^{\mathrm{\prime }})$ in the old basis. The reverse basis change also has integer coefficients, so $B\in {\mathrm{GL}}_r(\mathbb{Z})$. With these column coordinates, $A^{\mathrm{\prime }}=AB$; deleting row $k$ gives $A_k^{\mathrm{\prime }}=A_{k}B$ for every $k$.

The matrix $B$ of step 1.4 has an integer inverse, so [F7] makes $\det B$ a unit of $\mathbb{Z}$, and the description of the units of $\mathbb{Z}$ in [F7] gives $\det B=\pm 1$.

Independence of the deleted row: by step 1.2 and step 1.3 the matrix $A$ satisfies the hypotheses of [F6] with $m=r$, so for the determinants ${\mathrm{\Delta }}_k=\det A_k$ one has ${\mathrm{\Delta }}_k\mathrm{\ne }0$ and ${\mathrm{\Delta }}_k=(-1{)}^{k-1}{\mathrm{\Delta }}_1$ for every $k$; therefore $\mid \det A_k\mid =\mid {\mathrm{\Delta }}_1\mid$ is the same number for every deleted row, and it is positive because ${\mathrm{\Delta }}_1\mathrm{\ne }0$.

For every $k$, multiplicativity of the determinant [F7] applied to $A_k^{\mathrm{\prime }}=A_{k}B$ of step 1.4 gives $\det A_k^{\mathrm{\prime }}=\det A_k\det B$, so $\mid \det A_k^{\mathrm{\prime }}\mid =\mid \det A_k\mid \mid \det B\mid =\mid \det A_k\mid$ by step 2.1; hence every deleted-row determinant of the second system has the same absolute value as the first.

Combining steps 2.2 and 3.1: in the case $r\ge 1$ the number $R_K=\mid \det A_k\mid$ depends neither on the deleted row nor on the chosen system of fundamental units, and it is positive; in the case $r=0$ step 1.1 gives $R_K=1$. Therefore $R_K$ is well defined and is an invariant of $K$ alone, equal to $1$ in rank zero.

Choice accounting: the only place where AC enters is the existence of the systems of fundamental units in [F2], inherited from the AC-qualified unit theorem; the linear algebra of ranks, determinants and unimodular change of basis is elementary and choice-free.