Here is my attempt:
Since $x\geq 0$ and $x\in\mathbb{R}$, so there are two cases for $x$: $x > 1 $ or $ 0 \leq x \leq 1$
if $x > 1 $, then $x^2 < c < x\cdot c + x=x(c+1)\rightarrow x < c + 1$;
if $ 0 \leq x \leq 1$, then $x < c+1$. Thus set $S$ has a least upper bound, $\sup S$.
I am stuck at this step, can anyone give me a hit or suggestion to keep going.
(I am still learning the completeness axiom, haven't start Archimedian Property yet) Thanks.
The following proof doesn't need the Archimedean property:
For each $0\leqslant x<\sqrt c,$ define $$y:=\dfrac{x+\sqrt c}{2}.$$ Then $x<y<\sqrt c.$ This shows that given any $x\in S$ you will always find some $y\in S$ such that $y>x$ and so any upper bound of $S$ must be $\geqslant\sqrt c.$ Since clearly $\sqrt c$ is an upper bound of $S,$ we conclude that it is the least upper bound of $S,$ that is $\sqrt c=\sup S.$
Did you mean to say $x$ is in the Real numbers? If so:
For any $0 \le a < b$, then there is an $e = b - a > 0$ so $b = a + e$ and $b^2 = a^2 + 2\times a\times e + e^2 > a^2$.
So if $x \ge \sqrt{c}, x^2 >= c$ so $x$ isn't in $S$. So $\sqrt{c}$ is upper bound.
If $y < \sqrt{c}$ then $y^2 < c$ so $y$ is in $S$, so $\sqrt{c}$ is least upper bound.
HINT:Use the fact that $f(x)=x^2$ is an increasing and continuous function over an interval. And bijective for $x$ non-negative.
