According to Ballinger and Long (1959), the reaction mechanism is as such:
$\ce { (1) CH2ClCH2OH (aq) + OH^- (aq) -> CH2ClCH2O^- (aq) + H2O (aq)}$ (equilibrium)
$\ce { (2) CH2ClCH2O^- (aq) -> (CH2CH2)O (aq) + Cl^- (aq)}$ (slow)
This is not surprising since acid-base reactions, such as these proton transfers, take place very quickly. Although the second step may appear to be intramolecular, it may be slower since we are forming a strained epoxide ring and the $\ce {C-O^-}$ and the $\ce {C-Cl}$ bonds also need to be in a trans orientation for most effective orbital interaction of the relevant orbitals. Based on the above mechanism, your rate law seems to be correct, based on the assumption that the first step is much slower than the subsequent step and has come to establish an equilibrium.
Reference
Ballinger, P.; Long, F. A. The Reaction of Chlorohydrins and Hydroxide Ion in the Solvents H2O and D2O. J. Am. Chem. Soc., 1959, 81 (10), 2347–2352.