Given a vertex, find a path between it and each other vertex that minimizes the maximum weight in the path
Let $T = \{s\}$, and let $Q$ be a new empty priority queue. Insert each edge incident on $s$ in $Q$. While there are still nodes with $d$ from $s$ yet to be computed, extract and remove the minimum from $Q$. 

If such edge connects a node $u$ in $T$ with a node $v$ not in $T$, add each edge incident on $v$ that still isn't in $Q$ to $Q$, then set $M[v]$ to $\max\{M[u], w(u, v)\}$. 

Otherwise, disregard the extracted edge and extract more edges until you find a suitable one.

The overall complexity is $O(|E| \log |E|) = O(|E| \log |V|^2) = O(|E| \log |V|)$.

To informally prove correctness, we can observe that whenever we set $M[v]$ to a certain value $x$, we can certainly reach $v$ from $s$ with a path in which each edge weighs at most $x$. On the other hand, no better path exists, because we already gave each edge lighter than $x$ a chance to connect to $s$.