Let $d_{X,Y}$ be the weight assigned to the edge $XY$ of the graph. Clearly, in each triangle the maximum weight of an edge is the sum of the two others. Let $d_{A,B}$ be the maximum weight among all weights assigned to the edges. Now assign $0$ to vertex $A$ and $d_{A,X}$ to any other vertex $X$. Obviously, this labeling works for all edges including $A$. We claim that it works for all other edges.

Assume to the contrary that the labeling doesn't work for an edge $XY$ of the graph. So $d_{X,Y}$ is not the difference between $d_{A,X}$ and $d_{A,Y}$, which means $d_{X,Y} = d_{A,X} + d_{A,Y}$. Note that $B$ cannot be any of $X, Y$. Since $d_{A,B}$ is the maximum edge number, we have $d_{B,X} = d_{A,B} - d_{A,X}$ and $d_{B,Y} = d_{A,B} - d_{A,Y}$. Now consider triangle $BXY$. If $d_{X,Y}$ is the maximum weight of an edge in this triangle, we have
$$d_{X,Y} = d_{B,Y} + d_{B,X} \implies \begin{aligned}[t] d_{X,Y} &= d_{A,B} - d_{A,Y} + d_{A,B} - d_{A,X} \\ &= 2d_{X,Y} = 2d_{A,B}. \end{aligned}$$
Which is impossible since the weights are distinct. So the maximum edge weight in triangle $BXY$ is not $d_{X,Y}$. Without loss of generality, assume that $d_{B,Y}$ is the maximum one. So we have
$$d_{B,Y} = d_{B,X} + d_{X,Y} \implies \begin{aligned}[t] d_{A,B} - d_{A,Y} &= d_{A,B} - d_{A,X} + d_{A,X} + d_{A,Y} \\ &= d_{A,Y} = 0. \end{aligned}$$
Which is again a contradiction. This completes the proof. ■