Take any real closed field $K$ and denote its unique ordering by $\leq$. Prove if $f \in K[X]$ and $a,b \in K$ such that $a<b$ and $f(a) < 0 < f(b)$, then there is $c$ that $f(c)=0$.
I want use the fact that if we have real closed field $K$, so every polynomial $ f \in K[X]$ splits into irreducible factors of the form $X-a$ or $(X-a)^2 + b^2$ where $a,b \in K[X]$.
Let $f(x) = (x-d)((x-d)^2 + e^2))$ $ \ $ $d,e \in K$ and assume that for some $a,b \in K$: $a<b$, $f(a)f(b) < 0. $
$h(x)=(x-d)$ and $g(x)=(x-d)^2+e^2$
$sign(f)=sign(h)sing(g)$. Since $sign(g)=1$, so $sign(f)=sign(h)$. Assume that $sign(h(a))=sign(h(b)).$ It implies that $sign(f(a))=sign(f(b))$ which contradicts with $f(a)f(b)<0$, so we have $h(a)h(b)<0$.
Then $$h(a)<0<h(b)$$ $$a-d <0 <b-d$$ $$c:=d \in [a,b]$$ is a root of $f(x)$.
Is that correct?
Yes. You assume that $sign(h(a)) = sign(h(b))$. Then we have $sign(f(a)) = sign(f(b)) $, but we know that $ f(a) < f(b) $. Hence $ h(a)h(b) < 0 $ and we have
$$ \left. \begin{array}{l} h(a) =& a - d\\ h(b) =& b - d \end{array} \right\} \Longrightarrow a - d < b - d $$ Let $c := d$, then $$ a < c < b \quad \Longrightarrow \quad h(a) < h(c) = 0 < h(b) $$