= Solution
Retain the $(\lambda,\varepsilon)$-quasi-isometry $f:X\to T$ and the <Morse lemma for quasi-geodesics> constant $R$. Given $x,y\in X$, let $m$ be the midpoint of a geodesic $[x,y]$. There is a point $z$ on the tree geodesic $[f(x),f(y)]$ with
$$
d_T(f(m),z)\leq R.
$$
For any continuous path $\alpha$ from $x$ to $y$, choose a partition fine enough that consecutive $\alpha(t_i)$ are at distance at most one. Consecutive images under $f$ are then at distance at most
$$
D=\lambda+\varepsilon.
$$
Removing $z$ separates $f(x)$ from $f(y)$ in the tree along their geodesic, so this finite $D$-chain must contain an element within $D$ of $z$. For the corresponding $t_i$,
$$
d_T(f(m),f(\alpha(t_i)))\leq R+D.
$$
The lower quasi-isometry inequality yields
$$
d_X(m,\alpha(t_i))
\leq\lambda(R+D+\varepsilon).
$$
This constant depends only on the chosen quasi-isometry, so every <quasi-tree> has the <bottleneck property>.
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