Can Ethanol be used to carry out SN1 reaction? My book says that the reaction of

$$\ce{Cl-CH2-O-CH2-CH3}$$ with ethanol to form
$$\ce{CH3-CH2-O-CH2-O-CH2-CH3}$$
occurs through $\mathrm{S_N1}$ mechanism, but shouldn't it actually be $\mathrm{S_N2}$?

My understanding of this statement is that the substrate is a $1^\circ$ halide and on top of that ethanol is available to us as $\ce{C2H5-O-}$, which is a considerably strong nucleophile. All possible conditions are pointing towards an $\mathrm{S_N2}$ reaction but then why does the reaction proceed through $\mathrm{S_N1}$ mechanism?
Be careful!  Ethanol is not a strong acid ($\mathrm{p}K_\mathrm{a} \approx 16$). So its conversion to ethoxide anion is not so easy. Rather, it would attack the $\ce{C}$ center by its lone pair and then it would release proton to form the product.

Actually here what I think is, the hetero atom $\ce{O}$ just near to the $\ce{C}$ undergoing substitution will assist by its lone pair to kick out chloride and then being planar ethanol can attack from either side. And finally, you get the product of $\mathrm{S_N1}$ type.

But don't mess up between ethanol and ethoxide ion. They are totally different in nucleophilicity. Treat them as two different nucleophiles.