finding counterexamples to proofs.

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let f: A → B and let W, X ⊆ A.

Prove that if W ⊆ X, then g(W) ⊆ f(X)

I don't see how this can work, as I think I've found a counter example. Yet the instructions ask for a proof.

Let A = {0,1,2}

Let X = {0,1}

Let W = {0}

Then let f(x) = x+1 & let g(w) = w

Then g(0) = 0

f(0) = 1

f(1) = 2

So this satisfies all conditions, doesn't it? Yet, g(W) isn't a subset of f(X). Where have I gone wrong?