The set of all natural numbers is closed under addition

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I'm trying to prove the theorem described in the title, but my proof is so obvious I doubt it is sufficient.

Here's my way of proving it:

Definition of addition: Let $a$, $b$, and $c$ be natural numbers. We write $a + b = c$ if and only if there exist sets $A$ and $B$ such that a = equiv_class(A), b = equiv_class(B), $A \cap B = \emptyset$ and c = equiv_class($A \cup B$).

My Proof: Let $A, B$ be subsets of $\mathbb{N}$ (natural numbers set). Since both are subsets of $N$ then their union will also be a subset of $N$. Let $c$ belong to $C = A \cup B$. Now $C$ is a subset of $\mathbb{N}$, therefore $c$ is a natural number.

Is my proof so stupid it's funny?

What is the best way to prove this?