Show that the longest simple path from a node $x$ in a red-black tree to a descendent leaf has length at most twice that of the shortest simple path from node $x$ to a descendant leaf.
Both the shortest path $s$ and the longest path $l$ will have the same number of black nodes, as per property 5. Because of property 4, each red node must have a black parent and black children. This means that the number of red nodes must be less than or equal to the number of black nodes in any path in a valid red-black tree.
The biggest difference then, can be obtained if $s$ contains only black nodes and $l$ contains the same number of black nodes with the maximum possible red nodes added to it, which is $2s$.