Proving that any set X that has an upper bound and is non-empty, has a least upper bound

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so this is the problem I am having trouble with:

Let X ⊆ R be a set of real numbers that is non-empty and bounded from above. Then X has a least upper bound.

So, my approach was the following:

We know that X has an upper bound and let $y$ be the upper bound of $X$. If $y$ is the upper bound of $X$, then $y \geq x$ (for all $x \in X$).

So by contrapositive, this means:

If $y < x$, then $y$ is not an upper bound. Thus, $y$ is the least upper bound.

I would appreciate your help and any advice about my approach. Just to clarify, this is not part of any assignment, it is just an exercise I am doing to practice.