NMR coupling pattern for hydrocarbon
The two $\ce{CH2}$ groups are, as you mentioned, in chemically different environments. Thus, each one will couple separately. But within each of the two groups, there are two protons. So you will not get a doublet of doublets, but a triplet of triplets.

Note that for each $\ce{CH2}$ group the hydrogens are chemically equivalent but not magnetically. Therefore, if you include a chiral environment, the pattern will further complexify itself into a higher-order signal that cannot easily be analysed. But since you’re probably measuring in chloroform you need not worry — there is no chiral information present and the couplings will be degenerate.

If I had to label a spin system here, I would probably choose $\ce{A3BB'CC'MM'X3}$.

By the way, it can be assumed that both coupling constants to the two different $\ce{CH2}$ groups be reasonably similar that the overall pattern looks like a quintet without actually being one.