Suppose $f(x)$ is continuous on $[0,1]$ and $f(0) = 1, f(1) = 0$. Prove that there is a point $c$ in $(0, 1)$ such that $f(c) = c$.
let $l(x) = x$ and $d(x) = f(x) - l(x)$.
We have $d(1) = f(1) - l(1) = -1$ and $d(0) = 1$.
We divide $[-1, 1]$ in $H$ equal parts , where $H$ is a infinite hyperinterger.
$$-1, -1 + \delta, -1 + 2\delta, \ ... \ , -1 + H\delta = 1$$
Let $-1 + K\delta$ be the last partition point such that $d^*(-1 + K\delta) < 0$
$$\therefore d^*(-1 + K\delta) < 0 < d^*(-1 + (K+1)\delta)$$
$$\therefore f^*(-1 + K\delta) < -1 + K\delta < f^*(-1 + (K+1)\delta) - \delta$$
Since $f^*(-1 + K\delta) \approx f^*(-1 + (K+1)\delta) - \delta$, therefore $f^*(-1 + K\delta) \approx -1 + K\delta$
Let $c = st(-1 + K\delta)$
By taking standard part of the $f^*(-1 + K\delta) \approx -1 + K\delta$ , we get $f(c) = c$.
- I think this is probably correct but to be on the safe side please check my proof.
$\DeclareMathOperator{\st}{st}$This is indeed the correct and usual approach to proving the intermediate value theorem (and hence this question) under nonstandard analysis.
The fundamental idea is if $f\colon[a,b]\to\mathbb{R}$ is a continuous function and $u\in\mathbb{R}$ satisfies $f(a)<u<f(b)$ (or $f(b)<u<f(a)$) then there is a $c\in (a,b)$ such that $f(c)=u$. Why? $[a,b]$ has a finite subset $F$ containing its standard elements and therefore defining $x=\min\{x'\in F:u\leq f(x')\}$ gives $c=\st(x)$ provided $f$ was continuous. This is slightly more general than the idea you used where you provided a particular set $F$ (which in fact gives a simpler proof), but the idea is the same.