`Subsets` takes an optional 3rd argument as `Subsets[list, {n}, k]` that gives you the `k`th sublist of length `n`. Since your sublists are in sequence, you'll always need `k = 1`. You can then use this as:

 MapIndexed[First@Subsets[list, #2, 1] &, list]
 (* {{a}, {a, b}, {a, b, c}, {a, b, c, d}} *)

Another alternative would be:

 Reverse@Most@NestWhileList[Most, list, # != {} &]