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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. 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.