The chain metric is a metric, its topology is the weak topology, and the space is proper and complete

The length of a chain is independent of the chosen cells and is finite, so $d$ is a symmetric real-valued function with $d(x,x)=0$ and the triangle inequality. If $x_i,x_{i+1}$ lie in cells $C_p$ and $C_q$, then by the intersection condition of [F1] both points lie in $C_{p\wedge q}$, and the affine face isometries identify the three cells on their common points, so the Euclidean distances computed in $C_p$ and in $C_q$ agree; hence every choice gives the same sum. For finiteness, fix $x$ and let $A$ be the set of points joined to $x$ by a chain; if a cell $C_p$ meets $A$, say in $y$, then any $z\in C_p$ is joined to $x$ by the given chain followed by a one-step chain, so $C_p\subseteq A$: thus the trace of $A$ on every closed cell is either that cell or empty, and $A$ is open and closed in the weak topology of [F1]; as $X$ is connected, $A=X$. Therefore every two points are joined by a chain, each chain has finite length, and $0\le d(x,y)<\mathrm{\infty }$ for all $x,y$. Reversing a chain shows $d(x,y)=d(y,x)$, the one-term chain shows $d(x,x)=0$, and concatenating chains at $y$ and passing to the infimum shows the triangle inequality.

For every $x_0\in X$ and real $r>0$ the closed ball $\bar{B}(x_0,r)$ is contained in a finite union of closed stars. Call two vertices $v,w$ adjacent when they lie in a common cell; by [F2] each vertex has only finitely many adjacent vertices. Choose $v_0$ with $B(x_0,\delta )\subseteq \mathrm{st}(v_0)$ and let $S_r$ be the finite set of vertices reachable from $v_0$ by a walk of at most $K:=\lceil 2(r+1)\mathrm{/}\delta \rceil$ steps of the adjacency relation; then $\bar{B}(x_0,r)\subseteq {\bigcup }_{v\in S_r}\stackrel{}{\mathrm{st}}(v)$. Indeed, let $z\in \bar{B}(x_0,r)$. If $z=x_0$, it already lies in the starting star. Otherwise choose a chain $x_0=y_0,\dots ,y_m=z$ of length $0<\mathrm{\ell }<r+1$, delete consecutive repeated points, and traverse each remaining step at unit speed in its cell to obtain a map $\gamma \text{\hspace{0.17em}⁣}:[0,\mathrm{\ell }]\to X$ with $\gamma (0)=x_0$, $\gamma (\mathrm{\ell })=z$ and $d(\gamma (s),\gamma (s^{\mathrm{\prime }}))\le \mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }$, because the sub-chain between two parameter values has length at most the parameter difference. Put $h:=\delta \mathrm{/}2$ and let $0=s_0<s_1<\cdots <s_N=\mathrm{\ell }$ be the parameters obtained by steps of size $h$, the last gap being at most $h$; then $N\le K$. For each $j$ choose a vertex $v_j$ with $B(\gamma (s_j),\delta )\subseteq \mathrm{st}(v_j)$, taking $v_0$ for $j=0$. Since $d(\gamma (s_j),\gamma (s_{j+1}))\le h<\delta$, the point $\gamma (s_{j+1})$ lies both in $B(\gamma (s_j),\delta )\subseteq \mathrm{st}(v_j)$ and in $B(\gamma (s_{j+1}),\delta )\subseteq \mathrm{st}(v_{j+1})$, so the two open stars meet and $v_j,v_{j+1}$ lie in a common cell, that is, they are adjacent. By induction $v_N\in S_r$, and $z=\gamma (s_N)$ lies in $\mathrm{st}(v_N)\subseteq \stackrel{}{\mathrm{st}}(v_N)$.

$d$ is a metric: if $d(x,y)=0$ then $x=y$. Suppose $d(x,y)=0$. By the Lipschitz clause of [F2], $\mathrm{\mid }{\lambda }_v(x)-{\lambda }_v(y)\mathrm{\mid }\le Ld(x,y)=0$ for every vertex $v$, so ${\lambda }_v(x)={\lambda }_v(y)$ for all $v$. Since $x$ and $y$ correspond under the bijection $\mathrm{\Phi }$ of [F2] to the functions $\lambda (x),\lambda (y)$ on the vertex set (their values on the carrier), this gives ${\mathrm{\Phi }}^{-1}(x)={\mathrm{\Phi }}^{-1}(y)$ and hence $x=y$. With [step 1.1] this gives all the metric axioms of [F3].

Every metric ball is weakly open: the metric topology is contained in the weak topology of [F1]. Fix $x\in X$ and a cell $C_p$ with its Euclidean metric $d_p$. For $u,v\in C_p$ the reverse triangle inequality gives $\mathrm{\mid }d(x,u)-d(x,v)\mathrm{\mid }\le d(u,v)$, and the one-step chain gives $d(u,v)\le d_p(u,v)$; hence $d(x,\cdot )$ is $1$-Lipschitz, in particular continuous, on $C_p$. Therefore the trace of any open ball $B(x,r)$ on $C_p$ is relatively open, and as $p$ was arbitrary each ball is weakly open.

Every closed bounded subset of $X$ is compact. Let $A\subseteq X$ be closed and bounded; if $A=\mathrm{\varnothing }$ this is [F3]. Otherwise $A\subseteq B(x_0,r)\subseteq \bar{B}(x_0,r)$ for some $x_0$ and $r>0$. The ball $\bar{B}(x_0,r)$ is closed, since $y\mapsto d(x_0,y)$ is $1$-Lipschitz and hence continuous, and by [step 1.2] it is contained in the union of the finitely many closed stars $\stackrel{}{\mathrm{st}}(v)$, $v\in S_r$; each of those is compact in the weak topology by [F2] and therefore compact for the metric subspace topology by step 2.2: any metric-open cover is also weakly open on the star and thus has a finite subcover. Their finite union is metric compact, since an open cover has a finite subcover on each of the finitely many stars, and $\bar{B}(x_0,r)$, being a closed subset of that compact metric space, is compact by [F4]. Finally $A$, closed in $X$, is closed in the subspace $\bar{B}(x_0,r)$ and therefore compact by [F4].

The weak topology is contained in the metric topology. Let $W\subseteq X$ be weakly open and let $x\in W$. By [F2] choose a vertex $v$ with ${\lambda }_v(x)\ge 1\mathrm{/}(D+1)$ and $B(x,\delta )$ contained in the open star of $v$; the closed star $\bar{S}$ of $v$ is the image under $\mathrm{\Phi }$ of the realization $\mathrm{\mid }S\mathrm{\mid }$ of a finite subcomplex $S$ of $K$. The map $\mathrm{\Phi }{\mathrm{\mid }}_S\text{\hspace{0.17em}⁣}:\mathrm{\mid }S\mathrm{\mid }\to \bar{S}$ is a continuous bijection from the compact metric space $\mathrm{\mid }S\mathrm{\mid }$ of [F5] onto $\bar{S}$ with the weak topology, and the identity on $\bar{S}$ towards the metric subspace topology is continuous by [step 2.2], so the composite is a continuous bijection from a compact metric space onto a metric space; by [F4] the weak and metric topologies agree on $\bar{S}$. Hence $W\cap \bar{S}$ is open in the metric subspace $\bar{S}$: there is $\epsilon >0$ with $B(x,\epsilon )\cap \bar{S}\subseteq W$. Taking ${\epsilon }^{\mathrm{\prime }}:=\min \{\epsilon ,\delta \}>0$ and using $B(x,\delta )\subseteq \mathrm{st}(v)\subseteq \bar{S}$ we obtain $B(x,{\epsilon }^{\mathrm{\prime }})=B(x,{\epsilon }^{\mathrm{\prime }})\cap \bar{S}\subseteq W$. Therefore every weakly open set is metric open, and with [step 2.2] the two topologies coincide.

Every closed ball is compact by [step 3.1], since a closed ball is closed and bounded. For completeness, let $(x_k)$ be a Cauchy sequence in $X$; by [F3] fix an index $N$ with $d(x_m,x_k)<1$ for all $m,k\ge N$. The finitely many terms $x_0,\dots ,x_{N-1}$ have finite distances from $x_N$, so there is a real $\rho >0$ with $d(x_k,x_N)\le \rho$ for every $k$, and hence the sequence lies in the closed ball $\bar{B}(x_N,\rho )$, which is compact by [step 3.1] and therefore complete by [F4]; being Cauchy in $X$ and lying in this complete subspace, $(x_k)$ converges to a point of it, so $(X,d)$ is complete.