1. ## Set Proof

Is this a logical proof? (via one way containment method and proof by cases)

[Goal: Let A, B, C, and D be sets. If A⊆C and B⊆D then A∪B ⊆ C∪D]

By hypothesis, A⊆C ^ B⊆D
So x∈A→x∈C and x∈B→x∈D

(this is where I'm unsure. Am I allowed to state this?)
Let x∈ A∪B (by hypothesis?)
Then x∈A ∨ x∈B

(Is this correct?)
Since x∈A→x∈C and x∈B→x∈D
it follows that if x∈A ∨ x∈B then x∈C ∨ x∈D

By definition of a union, if A∪B then C∪D.
By definition of set containment, A∪B ⊆ C∪D.
Since A⊆C ^ B⊆D implies that A∪B ⊆ C∪D, it is proven

I really struggled with this one and still do not fully understand.
Is this using proof by cases? What is the hypothesis here?
Also, if this is right, what kind of improvements could I make as far as technicalities?

Thank you very much for you time. You are appreciated

2. Yes, your proof is correct.

(this is where I'm unsure. Am I allowed to state this?)
Let x∈ A∪B (by hypothesis?)
This is correct. By definition, A∪B ⊆ C∪D means that for every x, if x ∈ A∪B, then x ∈ C∪D. Any proof of the statement "For every x, if P(x) then Q(x)" starts with: "Fix any x and assume P(x)".

By definition of a union, if A∪B then C∪D.
It should say, if x ∈ A∪B, then x ∈ C∪D.