Euler-tour shortcutting of a doubled tree does not increase metric cost

Form the multigraph $M$ on $V$ whose edge list contains, for each tree edge $e=\{u,v\}\in E(T)$, exactly two parallel edges between $u$ and $v$ with length $d(u,v)$; these are nonloop edges because $u\mathrm{\ne }v$. Since $T$ is connected and spanning, so is $M$; each vertex degree is ${\mathrm{deg}}_M(v)=2{\mathrm{deg}}_T(v)$, because every nonloop edge contributes one to the degree of each endpoint, hence every degree of $M$ is even; and the total length of the edge list of $M$ is ${\sum }_{e\in E(T)}2d(e)=2w(T)$.

By [F4] the multigraph $M$ has an Euler circuit $C$, which by [F3] is a closed trail using every edge of $M$ exactly once, so its total length is the total length $2w(T)$ of the edge list of $M$. Since $n\ge 3$ and the spanning tree $T$ is connected, every vertex of $T$ has degree at least one, so every vertex of $V$ occurs on $C$.

Start at the first vertex $v_0$ of $C$ and list the vertices in order of their first visit, obtaining the distinct vertices $v_0=v_{i_1},v_{i_2},\dots ,v_{i_n}$ with $\{v_{i_1},\dots ,v_{i_n}\}=V$. The closed walk $C$ splits at these first visits into $n$ consecutive segments: for $k=1,\dots ,n-1$ the segment from $v_{i_k}$ to $v_{i_{k+1}}$, and the final segment from $v_{i_n}$ back to $v_{i_1}=v_0$. These segments partition the edges of $C$, so their lengths sum to $2w(T)$.

Replace each segment by the direct edge joining its two endpoints, which exists because the graph is complete; this gives the cyclic ordering $v_{i_1},v_{i_2},\dots ,v_{i_n}$ visiting every vertex exactly once, a feasible Hamiltonian tour. Iterating the triangle inequality along a segment bounds each shortcut edge by the length of that segment, since the metric is symmetric and all lengths are nonnegative; summing over the $n$ segments gives tour cost at most the total length of $C$, namely $2w(T)$.

The construction is explicit: the doubled multigraph is read off the finitely many tree edges, its Euler circuit is obtained by the constructive direction of [F4], and the first-visit scan of the closed walk takes time linear in its length, hence polynomial time in the encoded size of the instance. Therefore first-visit shortcutting of a doubled spanning tree yields a Hamiltonian tour of cost at most $2w(T)$.