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Fun problem. Let the weights of the edges be $w_1 \leq w_2 \leq \cdots \leq w_{n-1}$. For convenience, define $w_0=0$ and $w_1=1$. Let $e_1$, $e_2$, ..., $e_{n-1}$ be the edges, in the same order as the weights. Let $D$ be the matrix in the problem.

For $1 \leq k \leq n$, let $D_k$ be the matrix where $(D_k)_{ij}=1$ if $D_{ij} \geq w_k$ and $(D_k)_{ij}=0$ otherwise. Then $$D = \sum_{k=1}^n (w_k-w_{k-1}) D_k.$$ So it is enough to show that each $D_k$ is positive semidefinite. Moreover, it is enough to check this in the case that $w_k > w_{k-1}$, as otherwise the coefficient of $D_k$ is $0$.

So, assume $w_{k-1} < w_k$. We have $(D_k)_{ij}=1$ if and only if the path from $i$ to $j$ does NOT pass through edges $e_1$, $e_2$, ..., $e_{k-1}$. In other words, let $F_k$ be the forest obtained by deleting $e_1$, $e_2$, ..., $e_{k-1}$ from $T$; then $D_k$ is the block diagonal matrix whose blocks correspond to the connected components of $F_k$ and where each block is a square matrix entirely made up of $1$'s. This matrix is definitely positive semi-definite.

Moreover, if $w_{n-1} <1$, then $D_n$ is the identity matrix, so the sum is positive definite.

After writing this up, I checked the proof in Rivasseau and Wang (Theorem 2.2) and this is basically the same thing, so I don't know if you will like it any better.