It is perhaps simpler to say there is an x in H such that the index of x in H ≤ l.
Let the ideal H represent the inverse of an arbitrary member of the class group. (The entire class group is represented by ideals.) Select x as above, and let G be the ideal x/H. To contain is to divide, so G is indeed an ideal. Also, G and H are inverses in the class group. In other words, G represents the designated member of the class group.
Write GH = x and take indexes. The index of G is bounded by l.
Every element of the class group is represented by an ideal with index bounded by l. If the number of ideals with index ≤ l is finite, then the class group is finite.
Prime ideals generate the class group. Let P be such a generator. Other ideals that coincide with P (in the class group) are divisible by P. If P has order n, then an equivalent ideal H is P raised to an exponent that is 1 mod n, times other primes raised to multiples of their respective orders. Thus the index of H always exceeds the index of P. Find an equivalent H with index ≤ l, and the prime P has index ≤ l. The prime generators of the class group have index ≤ l. The order of such a generator is bounded by the log of l base |P|.