... there you are. The construction of the pseudometric is completely constructive. The cheating bit: you use Dependent Choice to create the sequence $\langle V_n:n\in\omega\rangle$. All you know is that for every neighbourhood $U$ of $e$ there is a neighbourhood $V$ of $e$ such that $V^3\subseteq U$ and you can't get the sequence without appeal to a higher authority; in this case DC is a high enough authority.

One more thought: you can cheat and avoid Urysohn's Lemma in Cartan's proof of his approximation theorem by taking a pseudometric $d$ on the group such that $B(e,1)\subseteq U$ and use it to create the Urysohn functions in the familiar way. To get the pseudometric take a sequence $\langle V_n:n\in\omega\rangle$ of symmetric neighbourhoods of $e$ such that $V_{n+1}^3\subseteq V_n$ for all $n$ and $V_0\subseteq U$; apply the proof of Kakutani's theorem that first-countable groups are metrizable to the sequence to get the pseudometric and $V_{n+1}\subseteq B(e,2^{-n})\subseteq V_n$ and ... .

My guess: Cartan was not (yet) aware of the use of AC in proving Urysohn's Lemma (the principle DC was formulated later and that Urysohn's Lemma implies a bit of choice was proved even later.

If you consider just the set of compact metric spaces and drop the isometry condition in the definition of the Gromov-Hausdorff metric all spaces have distance zero because you can make homeomorphic copies have arbitrarily small diameter in the Hilbert cube, and have these copies converge to the origin. You may want to try how you'd overcome this problem just in this particular case.

Alas, $b\mathbb{R}\setminus\mathbb{R}$ is not compact, so not homeomorphic to that torus.

First question: the map in the first half of the proof is continuous and injective on the remainder $b\mathbb{R}\setminus\mathbb{R}$, it maps that set onto $\{-1\}\times(b\mathbb{R}\setminus\mathbb{R})$. On the other hand: the closure of $f[\mathbb{R}]$ in the product is obtained by taking its union with the set $\{-1\}\times b\mathbb{R}$, so that map is not onto the remainder $B\mathbb{R}\setminus\mathbb{R}$.

Second question first: It appears that Theorem 4 is based (implicitly) on Theorem 2: the space $B\mathbb{R}$ has weight $\mathfrak{c}$ and hence is indeed embeddable into that torus, but the references to [4] indicate that the embedding Ivanov had in mind derives from the natural embedding of $b\mathbb{R}$ into that torus plus the homeomorphism from Theorem 2. So that proof is invalid as well, and your reference supports this.

That later remark does not say that ZFC+I implies global choice; it says that it provides a model for global choice. A well-order of $V_\kappa$ in the universe will feel like a global well-order if you think $V_\kappa$ is the whole universe.

We were working under the assumption that $X$ is $G$-pseudocompact, so if $f:X\to C(G)$ is equivariant and continuous then $f[X]$ has compact closure; because $X$ is dense in $\upsilon_GX$ the latter has this property too, so all (bounded) equivariant continuous maps from $\upsilon_GX$ can be extended to $\beta_GX$, that is, we have condition (b), because $\upsilon_GX$ is $G$-Hewitt condition (a) must fail, so $\upsilon_GX=\beta_GX$.

I recommend this paper by Jan de Vries, G-spaces: compactifications and pseudocompactness. The first part surveys methods of obtaining G-compactifications. The idea is to replace $\mathbb{R}$ by the space $C(G)$. When building $\beta_GX$ you use continuous functions $f:X\to C(G)$ whose image has compact closure. For building $\upsilon_GX$ you need a metric co-domain, so you look at the bounded functions only (finite distance from the zero-function); if you follow the product construction you do not necessarily get a compact closure.

Your definition may not require it but I think you can prove that you can identify $\upsilon_GX$ as a space between $X$ and $\beta_GX$: in Engelking's book $\upsilon X$ is constructed (Theorem 3.11.10) as a subspace of $\beta X$. Once you have done that my previous comment applies. And in that case $\upsilon_GX=\beta_GX$ because $\beta_G(\upsilon_GX)=\beta(\upsilon_GX)$ (as $\upsilon_G(X)$ is $G$-pseudocompact, because $X$ is) and because $X\subseteq\upsilon_GX\subseteq\beta X$ we have $\beta(\upsilon_GX)=\beta X=\beta_GX$. No apply the last sentence of my answer.

@MehmetOnat In the general case $\upsilon X$ sits between $X$ and $\beta X$; so you should establish that this also holds in the equivariant case. If $X$ is $G$-pseudocompact then so is $\upsilon_GX$ and as the latter is $G$-Hewitt you would be able to conclude $\upsilon_GX=\beta_GX$.

The definition does not work because of the "if and only if". If you weaken it to "The family of sets of the form $\{i\in I:x_i\in O\}$, where $O$ runs through the neighbourhoods of $x$, generates the ultrafilter $\mathcal{U}$" then the answer to question 1 remains the same but question 2 may get a more interesting answer.