Is this proof of the Archimedean Property valid?

Assume N subset R is bounded above.
So there is a least upper bound of N. Call it s.
If n in N, then n + 1 in N. Hence n + 1 <= s, n <= s - 1.
Thus s - 1 is a smaller upper bound of N, a contradiction.

Your proof works if and only if you have already proved

For every real number $s>0$, the set $\{ n\in\mathbb N \mid n<s\} $ contains a largest element.

I don't know exactly which axiomatic facts abour $\mathbb R$ (and its relation to $\mathbb N$) you have available, so it might be possible for you to prove this without already having the Archimedean property. But it doesn't really sound likely.

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(The proof suggested by William Elliot gets around this problem).