Prove that there is a positive number x such that $ x^3= 5$ 
Prove that there is a positive number $x$ such that $ x^3= 5$


Define $S=\{x\in\mathbb{R}\mid x>0,x^3<5\}$. 
Set $S$ is not empty because $1\in S$ and is bounded above because $2^3=8>5>x^3,\forall x\in S$, by the completeness axiom, $S$ has a least upper bound, $b$. 
Consider that if $b^3>5$, we can pick a small $\epsilon=(b^3-5)/(3b^2+3))$ such that $b-\epsilon<b$, then we can have $$(b-\epsilon)^3=b^3-3b^2\epsilon+3b\epsilon^2-\epsilon^3>b^3-3b^2\epsilon-3\epsilon^3>b^3-3b^2\epsilon-3\epsilon=5\tag 1$$
So $(b-\epsilon)^3>5>x^3,\forall x\in S$ which contradicts $b$ is the least upper bound. 
Now consider that $b^3<5$, then so we can pick a $\epsilon=(5-b^3)/(3b^2+3b+1)$ such that $b+r>b$. Then we have $$(b+r)^3=b^3+3b^2\epsilon+3b\epsilon^2+\epsilon^3<b^3+3b^2\epsilon+3b\epsilon+\epsilon=5 \tag 2$$ 
This shows $b$ is in $S$ which contradicts $b$ is an upper bound. Thus, by the positivity axiom, $b^3=5$.

I am following the direction provided from the book, but I am not really good at inequity, so I am not sure $(1)$ and $(2)$ correct or not. If not, can someone give me a hit or suggestion to make $(1)$ and $(2)$ correct? Thanks in advanced. 
 A: The first part of your proof looks good! Nice choice of $\varepsilon.$ That establishes that $S$ has a least upper bound $b$ and $(1)$ shows that $b \leq 5$.
There are some bumps on the second half but you are definitely on the right track. First, to be pedantic, looks like you used $r$ on two occasions instead of $\varepsilon$. No biggie. The real problem is with your reasoning about $b$. Note that $b$ is still acting as the least upper bound for $S$, and it is perfectly fine for a least upper bound of a set to be an element of the set itself. For example, consider the closed unit interval $[0,1]$. It's least upper bound is $1$, a member of $[0,1]$.
That being said, $(2)$ was executed excellently and still completes the proof. The reasoning is that $(2)$ reveals a number $b+\varepsilon \in \Bbb{R}$ such that $b^3<(b+\varepsilon)^3<5$. Hence $b+\varepsilon \in S$, so the contradiction is that you've found an element in $S$ larger than the least upper bound of $S$. 
P.S. You ask great questions. Complete with $\LaTeX$, a clear summary of what you've tried so far and you respond to input from other MSE users. Keep it up!
