Nested Interval Property implies Axiom of Completeness

You need to make the gap between the upper bounds and $S$ shrink to $0$.

HINT: Given $s_n\in S$ and an upper bound $K_n$ of $S$, let $x_n=\frac12(s_n+K_n)$. If $x_n$ is an upper bound for $S$, let $s_{n+1}=s_n$ and $K_{n+1}=x_n$. If not, choose $s_{n+1}\in[x_n,K_n]$, and let $K_{n+1}=K_n$. Now consider the intervals $I_n=[s_n,K_n]$.

Actually Nested Interval Theorem implies Completeness Axiom, only if you assume that Archimedean Property holds. Under this hypothesis, as Brian M. Scott mentioned earlier, you can make the gap between the upper bounds and S shrink to 0.

Completeness Axiom implies both Archimedean Property and Nested Interval Theorem!!!! Though in Non-Archimedean Ordered Fields (like the field of Rational Polynomials) the Nested Interval Property ALONE does not entails Completeness Axiom

So in a general setting, Completeness Axiom is equivalent to Nested Interval Property+Archimedean Property

It's not valid in its current state: All you can conclude is that $M$ is an upper bound for $S$. What happens, say, if we choose $S = [0, 1]$ and a sequence

$$K_n = 2 + \frac 1 n$$

If our initial choice of $s$ is $0$, then our intervals are $$I_n = \left[0, 2 + \frac 1 n\right] \implies \bigcap_n I_n = [0, 2]$$

So $M = 2$.

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The gap starts in the line beginning "Also $M$ is the ...."