Prove that any non-negative real number $a$ has $n$th root.
By definition an $n$th root of a number $x$ is a number $r$ such that $r^n=x$.
Consider the set $\mathcal B=\left\{x: 0\le x \le 1+a\right\}$,then $(\mathcal B, \le)$ is a partially ordered set containing minimum and maximum element,define a function $f: \mathcal B \to \mathcal B$ with $$ x \mapsto x+ \frac{a-x^n}{n(1+a)^{n-1}}$$
Such function is order-preserving since for $0 \le x \le y \le 1+a$ we have that:
$$f(x)-f(y)=(x-y)(1-\frac{\sum_{k=0}^{n-1}x^ky^{n-1-k}}{n(1+a)^{n-1}})$$
Besides we know that $0 \le c \le d \implies c^n \le d^n$ for $n \in \mathbb N$,using this we have that $$0 \le \sum_{k=0}^{n-1}x^ky^{n-1-k} \le n(1+a)^{n-1}$$
Which implies that $f(x) \le f(y)$,using Knaster–Tarski theorem follows that exists $r$ in $\mathcal B$ such that : $$r+ \frac{a-r^n}{n(1+a)^{n-1}}=f(r)=r$$
And from the definition of $n$th root the desired result does hold.
I like to know if my proof is right and if it's not true then I want to know the errors.
Your proof is correct! Perhaps, a less-involved approach (or another alternative) is as follows; first $0^n=0$, $1^n=1$, and $a^{1}=a$, so we may assume that $a\notin\{0,1\}$ and $n\ge 2$. Now, consider the continuous function $$f\colon [0,a]\to\mathbb{R}\,,~\,~\,~\,~ x\mapsto \left\lbrace \begin{array}{ll} x^n-a &\mbox{if $a>1$}\\ a-x^n &\mbox{if $a<1$} \end{array}\,. \right. $$ It follows that $f(0)$ and $f(a)$ are non-zero and of opposite signs; hence, by the Intermediate Value Theorem, there exists $r\in(0,a)$ such that $f(r)=0$——that is, $a=r^n$.