I don't think your argument quite works.  In particular, you can't really conclude anything about the lengths of your paths.  Here's a sketch of a proof using [Menger's Theorem](http://en.wikipedia.org/wiki/Menger%27s_theorem).  By way of contradiction, assume that the edge-connectivity of $G$ is 5. Let $u, v \in V(G)$.  By the edge-version of Menger's theorem, there are 5 edge-disjoint paths between $u$ and $v$.  Since, $G$ is 5-regular, no three of these paths can intersect at a common vertex (other than $u$ or $v$).  It is thus easy to construct 3 vertex disjoint paths between $u$ and $v$.  By the vertex-version of Menger's theorem, $G$ is 3-connected, which is a contradiction.