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Every path in a minimum spanning tree solves the bottleneck shortest path problem between its endpoints. If $e$ is a maximum-weight edge on the tree path, deleting it gives a graph cut separating the endpoints. A competing path with every edge lighter than $e$ would contain a crossing edge $f$; replacing $e$ with $f$ would decrease the tree's total weight, a contradiction. Equal weights cause no problem: the argument excludes a strictly smaller bottleneck and does not require a unique tree.

Take a minimum spanning tree $T$, for example one constructed by Kruskal's algorithm. Fix $v$ and let $e$ be a longest edge of the tree's unique $xv$ path. Deleting $e$ partitions $T$ into two components separating $x$ and $v$. Any alternative $xv$ path crosses this graph cut. If all of its edges were shorter than $e$, a crossing edge $f$ would let us replace $e$ by $f$, producing a spanning tree of smaller total length. This contradicts minimality of $T$.

Therefore every competing path has an edge at least as long as $e$, and the tree path solves the bottleneck shortest path problem. The same minimum spanning tree works simultaneously for every $v$; this is the minimax path property of a minimum spanning tree.

Let $b_{\ast }$ be the smallest threshold for which the edges of length at most $b_{\ast }$ connect $x$ to $y$. This is exactly the optimum value of the bottleneck shortest path problem. While any edge longer than $b_{\ast }$ is being processed, every edge of some threshold-$b_{\ast }$ route is still present, because shorter edges have not yet been processed. Thus all edges longer than $b_{\ast }$ are deleted. The final path from part (i) consequently has maximum edge length at most $b_{\ast }$. It cannot have a smaller maximum by the definition of $b_{\ast }$, proving optimality. Ties in edge ordering do not affect this argument.