Proving NP-Complete problem by reduction of subset-sum
To prove a problem is NP-complete you have to prove it is in NP and that it is NP-hard.
NP-hardness proof is usually however, a reduction from one NP-hard problem. To prove that the problem is NP, you can either describe  a non-deterministic polynomial time algorithm for the problem (An NTM that decides the language of the yes-Instances in polynomial time), or you can show that there is a certificate of polynomial size for each yes-instance, such that given the instance with the certificate you can in polynomial time prove that the instance is a yes instance.

Both ways are quite easy here. A polynomial certificate would be an answer of the problem. You can verify if a given set of words as a solution really uses each letter exactly once. Describing an NTM is also straightforward, where the NTM guesses all possible subsets of words and checks for each subset if it covers all letters and each letter exactly once.

To prove NP-hardness, I suggest a reduction from the Exact-Cover problem instead, since it is more related. In the exact cover problem, you are given a Universe of elements $\mathcal{U}$ and a family of sets $\mathcal{F} = \{F_1, \dots F_n\}$ over the universe (A set of subsets of the universe). The output is whether there is a subset of the sets in the family that cover the whole universe and each element is covered exactly once (included in exactly one chosen set). A reduction to your problem would be as follows. We set $\Sigma = W = \mathcal{U}$ (each element in $W$ is repeated exactly once). From each set $F \in \mathcal{F}$ we build a word $s$ that contains exactly the elements in the set (in any order).

If the instance of the exact-cover problem is a yes-instance, then there is a subset of the family $\{F_{i_1}, \dots F_{i_k}\}$ that covers each element in the universe exactly ones. The words $s_{i_1}, \dots s_{i_k}$ we built from these sets cover each element in $W$ exactly once so it is also a yes instance. On the other hand, if the instance of your problem is a yes instance, there is a ste of words $s_{i_1}, \dots s_{i_k}$ that covers each element in $W$ and hence in $\Sigma$ exactly once. The sets $F_{i_1} \dots F_{i_k}$ we build the words from cover each element in the universe exactly once.