This paper by P.-E. Caprace, uses a "refined boundary" of a CAT(0) space. This boundary is constructed in the following way : given a point $\xi$ in the boundary at infinity of your space $X$, you construct a point $X_\xi$, which is the inverse limit of the horoballs centered at $\xi$. Here the maps $\phi_{m,n}$ are the CAT(0) projections. Then the space $X_\xi$ is itself CAT(0), and you can iterate the construction. Under reasonable hypotheses, the construction stops after a finite number of steps, and the refined boundary is the union of all the spaces you get.
(In the case of symmetric spaces, this construction has been already considered by Karpelevic in 1965, but with different definitions, and I don't think he saw it as inverse limits).
I can complete Anton's plan with an additional assumption that geodesics do not branch. I also assume local compactness (otherwise there are too many technical details to deal with). More precisely, I prove the following:
Let $X$ be a geodesic space. Suppose that
For every finite subset $Q\subset X$, every similarity $Q\to X$ can be extended to a bijective similarity $X\to X$. A similarity is a map that multiplies all distances by a constant. (It is easy to see that, in the case of a length metric, the more complicated definition from the question reduces to this.)
$X$ is locally compact.
For every two distinct points of $X$ there is a unique line containing them. (A line is a subset isometric to $\mathbb R$. The existence of lines follows from the two other assumptions.)
Then $X$ is isometric to $\mathbb R^n$ for some $n$.
Proof. First observe that the group of similarities acts transitively on pairs (oriented line, point not on this line). Indeed, given two such pairs $(\ell_1,p_1)$ and $(\ell_2,p_2)$, it suffices to find similar triples $p_1,x_1,y_1$ and $p_2,x_2,y_2$ where $x_i,y_i$ is a positively oriented pair of points on $\ell_i$. And it is easy to see that such triples $p_i,x_i,y_i$ realise all similarity types of, e.g. isosceles triangles.
Next we construct perpendiculars. Lines $\alpha$ and $\beta$ intersecting at a point $q$ are said to be perpendicular if there exists an isometry that fixes $\alpha$ and maps $\beta$ to itself by reflection in $q$. It is easy to see that this relation is symmetric. Further, for every line $\ell$ and every point $p\notin\ell$ there exists a unique perpendicular to $\ell$ containing $p$. To prove existence, pick two points $x,y\in\ell$ such that $|px|=|py|$. There is an isometry that fixes $p$ and exchanges $x$ and $y$. It acts on $\ell$ by reflection in the midpoint $q$ of $[xy]$, hence the line $(pq)$ is perpendicular to $\ell$. Uniqueness follows from the fact that two distinct reflections of $\ell$ cannot fix $p$, otherwise their composition shifts $\ell$ along itself and fixes $p$, and some iteration of this shift would break the triangle inequality.
This argument also shows that the base $q$ of the perpendicular is the nearest point to $p$ on $\ell$, and the distance from $p$ to $x\in\ell$ grows monotonically with $|qx|$.
Clearly all right-angled triangles with given leg lengths $a$ and $b$ are isometric; denote their hypotenuse by $f(a,b)$. Then $f$ is strictly monotone in each argument and positively homogeneous: $f(ta,tb)=tf(a,b)$.
Next, we show that the sum of Busemann function of two opposite rays is zero. Or, equivalently, if $\gamma$ is an arc-length parametrized line, $q=\gamma(0)$ and a line $(pq)$ is perpendicular to $\gamma$, then $B_\gamma(p)=0$ where $B_\gamma$ denotes the Busemann function of $\gamma$. Suppose the contrary. We may assume that $|pq|=1$ and $B_\gamma(p)=c>0$. This means that for every point $x\in\gamma$ we have $|px|-|qx|\ge c$. Let $s\in\ell$ be very far away and let $p_1$ be the perpendicular from $q$ to $(ps)$. The above inequality implies that $|pp_1|\ge c$, hence we have an upper bound $|qp_1|\le\lambda<1$ where $\lambda$ is determined by $f$ and $c$ and does not depend on $s$. Let $q_1$ be the base of the perpendicular from $p_1$ to $(qs)$, then rescaling the above inequality yields that $|p_1q_1|\le\lambda|qp_1|\le\lambda^2$. The next perpendicular (from $q_1$ to $p_2$ on $(ps)$) has length at most $\lambda^3$ and so on. Summing up these perpendiculars, we see that $|ps|\le 1/(1-\lambda)$ which is not that far away, a contradiction.
The fact that opposite rays yield opposite Busemann functions implies that the Busemann function of a line $\gamma$ is the only 1-Lipschitz function $f$ such that $f(\gamma(t))=-t$ for all $t$. And, given local compactness and non-branching of geodesics, this implies that $X$ is split into lines parallel to $\ell$, where a line $\gamma_1$ is said to be parallel to $\gamma$ if the Busemann function of $\ell$ decays with unit rate along $\gamma_1$ (to construct a parallel line, just glue together two opposite asymptotic rays). Further, Busemann functions of parallel lines coincide, due to their above mentioned uniqueness.
Now it is easy to see that the level set of a Busemann function (which is also a union of perpendiculars to a given line at a given point) has the same isometry extension property and is a geodesic space. Then we can carry induction in the parameter $d$ defined as the maximum number of pairwise perpendicular lines that can go through one point. The cases $d=1$ and $d=2$ can be done by hand, then use isometries exchanging perpendicular lines to do the induction step.
Best Answer
The metric category of metric spaces and metric maps (i.e. Lipschitz with constant $1$, i.e. never stretching) has a certain rigidity to it so that the idea of a metric base seems attractive. However, I feel that the metric dimension makes sense only for a somewhat limited class of, rather than for all, metric spaces. (Possibly, more than one of several competing definitions can make sense).
Consider (see the Question above):
Property G A subset $B$ of a metric space $(M,d)$ is called a metric g-set for $M$ if and only if $$ \forall_{b \in B}\ d(x,b)=d(y,b)\ \ \implies\ \ x = y $$
Every such set $B$ was called a metric basis above. Let $B$ be one, and let $\ B\subseteq C\subseteq M.\ $ Then $$ \forall_{c \in C}\ d(x,c)=d(y,c)\ \ \implies\ \ x = y $$
Thus, $\ C\ $ has the metric g-set property too. We see that declaring Property G as a definition of metric basis goes against the spirit of basis in general. Instead, Property G corresponds to the general notion of a generating set.
It took quite a long time to arrive at the topological definition of a curve and of the topological dimension (from Cantor's curves in $\ \mathbb R^2,\ $ to Brouwers-Urysohn-Menger, then to the algebraic topological dimensions, etc.).
It took also considerable time to arrive at a general definition of basis which ultimately led to matroids.
In this thread, the topic is rather something like the metric variations of the independent sets and of a basis. Perhaps one should eye matroids for a comparison. These days things go much faster than in the past but the process of obtaining one or more of such sound metric notions may still take some tries and patience.
Thus let's examine the property of a basis, and more generally, of independent sets for matroids: let $X$ be a matroid; then
if $\ A\subseteq X\ $ is an independent set, and $\ b\in X,\ $ then there exists $\ a\in A\ $ such that $\ (A\setminus\{a\})\cup\{b\}\ $ is independent;
if $\ A\subseteq X\ $ is an independent set (resp. basis), and $\ b\in X\ $ depends on $\ A\ $ (resp. $\ b\in X\ $ is arbitrary), then there exists $\ a\in A\ $ such that $\ (A\setminus\{a\})\cup\{b\}\ $ generates the same set as $\ A\ $ (resp. $\ (A\setminus\{a\})\cup\{b\}\ $ is a basis).
Now let's look at some metric spaces.
EXAMPLE 1 Let $\ X:=\{0\ 1\ 2\ 3\},\ $ and let $\ d\ $ be a metric (thus, symmetric, etc) in $\ X,\ $ such that: \begin{eqnarray} d(0\ n) &:=& 2-\frac 1n&\qquad \mbox{for }\ \ n=1\ 2\ 3\\ d(k\ n) &:=& 2&\qquad \mbox{for every}\ \ 0<k<n\le3 \end{eqnarray} We see that there is a $1$-element basis $\ \{0\}\ $ in $\ (X\ d)\ $ in the sense of Question while there is no other $1$-element basis. However, in the spirit of matroids (see the above properties 1. and 2.), every $1$-element subset of $X$ should be a basis. Well, this is not so.
ALSO: Every $2$-element set $\ X\subset\{0\ 1\ 2\}\ $ is a minimal G-set (basis). These three $2$-element sets and $\ \{0\}\ $ are the only minimal G-sets; thus, there are four of them altogether.
Turning to continuous spaces will not help:
EXAMPLE 2 Let $\ X:=[1;\infty)\ $ be a closed half-line, and let $\ d\ $ be a metric (thus, symmetric, etc) in $\ X,\ $ such that: \begin{eqnarray} d(1\ x) &:=& 2-\frac 1x&\qquad \mbox{for every}\ \ x > 1\\ d(x\ y) &:=& 2&\qquad \mbox{for every}\ \ 1<x<y \end{eqnarray} We see that there is a $1$-element basis $\ \{1\}\ $ in $\ (X\ d)\ $ in the sense of Question while there is no other $1$-element basis. Again, in the spirit of matroids, every $1$-element subset of $X$ should be a basis.
Thus, continuity didn't help.
ALSO: In addition to $\ \{0\},\ $ the only other minimal G-sets are sets $\ X\subset[0;\infty)\ $ such that $\ |[0;\infty)\setminus X|=1$.
CONCLUSION If (a big if) we had to continue with the definition of basis from Question then we would have to restrict the class of the metric spaces under consideration. The first candidate which comes to mind would be the class of transitive space (i.e. such that for every two points there is an isometry of the space onto itself, which sends one of the points onto the other one).