In step 2, it's shown that x ∈ A ∩ B, so in step 3, it follows by the definition of set intersection that x ∈ A and x ∈ B.
Then in step 4, by definition of image of an element, both F(x) ∈ F(A) and F(x) ∈ F(B).
The rest of the proof follows.
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