 Solution

Choose a root $o\in T$ and finite sets $F_n\subset T$ whose union is dense, arranging that $F_n$ is a $2^{-n}$-net and $F_n\subset F_{n+1}$. Let $T_n$ be the finite subtree spanned by $o$ and $F_n$. A depth-first contour traversal of $T_n$, recording distance from $o$, gives a continuous excursion $g_n$ whose real tree encoded by an excursion is $T_n$.

The traversals may be chosen compatibly: when passing from $T_n$ to $T_{n+1}$, insert the new branch traversals into small time intervals at their attachment points. Since every new component has height at most $2^{-n+1}$, choose the time changes so thatAfter harmlessly taking a faster sequence of nets, these errors are summable. Hence $(g_n)$ is uniformly Cauchy and converges uniformly to a continuous $g:[0,1]\to {\mathbb{R}}_{+}$ with $g(0)=g(1)=0$.

The net property gives $d_{GH}(T_n,T)\to 0$. By the stated continuity of excursion coding, $d_{GH}(T_{g_n},T_g)\to 0$. Since $T_{g_n}$ is isometric to $T_n$, uniqueness of limits in the Gromov-Hausdorff distance implies that $T_g$ is isometric to $T$. This proves the excursion coding theorem for compact real trees.