How is it possible, that the hydrolysis of ethers proceed via a SN2 mechanism?
While the mechanism may look simple in the way you described it there, it doesn’t work exactly like that.

First, you are in the presence of (diluted) $\ce{H2SO4}$ so aside from extremely short-lived accidental self-deprotonation of water, $\ce{OH-}$ ions will not be around. Therefore, $\ce{OH-}$ cannot act as an attacking group, whatever the mechanism.

Second, an alcoholate is one of the worst leaving groups in $\mathrm{S_N}$ reactions one can imagine. The only case where you will frequently see leaving alcoholates is for epoxide openings. But that’s not a problem here, we have $\ce{H2SO4}$. We can protonate the ether oxygen so that the leaving group will now be an alcohol.

Third, since $\ce{OH-}$ cannot be the attacking species, we need to choose from the nucleophiles present. Either $\ce{HSO4-/SO4^2-}$ or $\ce{H2O}$ can attack the $\alpha$ carbon (in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ manner). In the former case, a later attack of $\ce{H2O}$ to liberate the sulphate needs to be assumed. Both attacks work, but neither nucleophile is strong, so the driving force must be the liberated alcohol.

Fourth, the attacking $\ce{H2O}$ will create a protonated alcohol. This needs to deprotonate to give the free alcohol. Thereby the proton we used in second (step 1 of the mechanism, if you wish) is reliberated and therefore the reaction is acid-catalysed.

Finally, one needs to consider the possibility of equilibrium. The reaction can proceed in both directions. Usually by the choice of solvent one can direct it into one or the other. Hydrolysis is usually achieved with a large surplus of water, driving the reaction towards the alcohols by Le Chatelier’s principle. The reverse reaction would be triggered by removal of water or high concentrations of either alcohol.