Hello. I want make sure im doing right step.

Prove the supremum of the set A={2-1/n: n in natural number} is 3.

My answer
Proof: lets prove supA=2
Then 2 is an upper bound of A.
Let S'<2, then S' is not upper bound.
There exists x in A such that x> S'
Then 0<2-S'
There exists n in natural number such that 0<1/n <2-S'
Therefore supA=2. QED.

I mean prove the supremum of the set {2-1/n: n in N} is 2

Your answer is mostly correct, but there is a small mistake. You need to prove that the supremum of the set A is not 2, but actually 3. Here's how you can correct it:

Proof: Let's prove supA = 3.
First, we need to show that 3 is an upper bound of A.
For any x in A, we have x = 2 - 1/n, where n is a natural number.
Since n is positive, it follows that 1/n is positive and less than or equal to 1. So, 2 - 1/n is always less than or equal to 2.
Therefore, for any x in A, we have x ≤ 2.

Now, we need to show that 3 is the least upper bound, or supremum, of A.
Assume, for the sake of contradiction, that there exists an upper bound S' of A such that S' < 3.
Since S' is an upper bound of A, we have x ≤ S' for any x in A.
Choose n = 1/(3 - S'). This is a positive natural number, since S' < 3.
Then, we have 0 < 1/n = 3 - S'.
Rearranging this inequality, we get S' < 3 - 1/n = 2 - 1/n + 1 = x + 1, where x = 2 - 1/n is an element of A.
But this contradicts the assumption that S' is an upper bound of A.
Therefore, no upper bound S' of A can be less than 3.

Hence, we have shown that 3 is an upper bound of A, and there is no upper bound of A less than 3.
Therefore, supA = 3. QED.

By following the corrected steps, you can successfully prove that the supremum of the set A = {2 - 1/n : n is a natural number} is indeed 3.