An $\mathrm{S_N2}$ reaction is a bimolecular reaction. As such, it requires both partners to be sufficiently reactive; that is we need both a good-enough nucleophile to attack and a good-enough leaving group. If we have a slightly better leaving group, we can make do with a worse nucleophile and vice-versa, but overall they have to be strong enough in combination.
Leaving groups in the $\mathrm{S_N2}$ reaction are good ones if they have a good charge distribution (charge per volume), which correlates well with being the conjugate base of a strong acid. If your leaving group is cationic (i.e. becomes neutral after leaving), it is a very good leaving group as there is no charge that needs to be distributed. Hence why protonating a hydroxy group to give $\ce{-OH2+}$ turns it into a good leaving group.
This is also why bromide is a rather good leaving group: Not only is it the conjugate base of a very strong acid $\ce{HBr}$, it is also a very large anion with a small charge and thus a rather good charge distribution. On the other hand, a hydroxide, which would be the leaving group in an $\mathrm{S_N2}$ attack onto an alcohol, is much smaller but still has the same charge. Thus, it is an absolutely terrible leaving group and nucleophilic attacks that liberate hydroxide effectively don’t happen.
Thankfully for synthetic organic chemistry, there are many ways of turning hydroxy groups into good leaving groups; most of these involve adding an acyl residue of some sort that creates a stabilised anion upon leaving with a much better charge distribution. The most popular ones (given as the actual leaving group) are probably mesylate $\ce{MsO-}$, tosylate $\ce{TsO-}$ and triflate $\ce{TfO-}$ — all sulphonic acid derivatives.
As a side note: You wrote an addition of $\ce{Na+}$ to the hydroxy group to generate a charged species. But unlike a proton — which is such a strong Lewis acid that it will never walk alone — a sodium cation will never form that strong a bond with a hydroxy group that they would leave as $\ce{NaOH}$. In fact, $\ce{NaOH}$ if it doesn’t precipitate due to low solubility will always dissociate into $\ce{Na+}$ and $\ce{OH-}$ while water molecules will not in typical organic solvents.