Djikstra algorithm analysis
8.    while (the priority queue is not empty):
 9.        u ← delete-min from priority queue

10.        For each edge (u,v) ∈ E leaving u:
11.            If d(v) &gt; d(u) + l(u,v):
12.                decrease-key of v to d(u) + l(u,v) in priority queue
13.                d(v) ← d(u) + l(u,v)


Consider the multi-set of all edges that appear on line 10, where $u$ goes through all vertices thanks to the line 9 inside the while loop. There are two cases. 


The graph is undirected. Each of those edges must be incident to a vertex $u$. So each edge in the graph appears exactly twice.
The graph is directed. Each of those edges must have $u$ as its starting endpoint. So each edge in the graph exactly appears once.


So in total, the number of edges that appear on line 10 is either $2|E|$ with undirected graph or $|E|$ with directed graph. So the running time from lines 8, 10, 11, and 12 is $2mO(\log(n))$ or $mO(\log(n))$. That is, $O(m\log(n))$.

Had the line 10 been For each edge (w,v) ∈ E:, your analysis might have had a chance to be correct.