Show the IVP of the zeros of a family of polynomials with continuous coefficients A quite well known result in complex analysis is that "the zeros of a polynomial depend continuously on its coefficients". More precisely (as formulated by Mishael Zedek (1964) in Theorem I):

Theorem. Given a polynomial $p(s) = p_ns^n + p_{n-1}s^{n-1} + \cdots + p_0$,
  $p_n \neq 0$, an integer $m \geq n$ and a number $\epsilon > 0$, there
  exists a number $\delta > 0$ such that whenever the $m+1$ complex
  numbers $q_k$, $0 \leq k \leq m$, satisfy the inequalities
       $$ |q_k - p_k| < \delta \text{ for } 0 \leq k \leq n \text{ and, if } m > k, |q_k| < \delta \text{ for } n+1 \leq k \leq m $$
then the zeros $\mu_k$, $1 \leq k \leq m$, of the polynomial $q(s) =
> q_ms^n + q_{m-1}s^{n-1} + \cdots + q_0$ can be labeled in such a way
  as to satisfy with respect to the zeros $\lambda_k$, $1 \leq k \leq
> n$, of $p(s)$ the inequalities
$$ |\mu_k - \lambda_k| < \epsilon \text{ for } 1 \leq k \leq n \text{
> and, if } m > k, |\mu_k| > 1/\epsilon \text{ for } n+1 \leq k \leq m
> $$

But now consider the following lemma:

Lemma. For any family of $n$-th degree nonconstant polynomials $p(s, \xi) = p_0(\xi)s^n +p_1(\xi)s^{n-1} + \ldots + p_n(\xi)$, $\xi \in [a, b]$, $p_i \subset C[a, b]$ and $p_0(\xi) \neq 0$, the real part functions of its zeros have the intermediate value property (IVP).

With the statement that every zero's real part function has the IVP we understand that we can factorize $p(s, \xi)$ into a product $p(s) = c(\xi)(s - \lambda_1(\xi))(s - \lambda_2(\xi)) \ldots (s - \lambda_n(\xi))$ such that every $\Re(\lambda_i(\xi)) : [a, b] \to \mathbb{R}$ ($i \in \{1, 2, \dots, n\}$) has the IVP.
I constructed this lemma in order to give a proof for the Routh-Hurwitz theorem based on a continuity argument. It is not homework.
Obviously the thought goes like: prove that we can build (gather the correct zeros together) $\lambda_i(\xi)$ such that $\Re(\lambda_i(\xi))$ is continuous on $\xi \in [a, b]$, then use the intermediate value theorem to establish the IVP. But is this lemma true? If so, could you hint me to a proof? Your help is very much appreciated!
 A: The obvious (but surprisingly difficult) way to do this is to use the theorem of Zedek that you cite to construct continuous function $\lambda_i$ following the roots of the polynomial. Essentially, our strategy will be to figure out some way to label the roots of the polynomial - and we'll do that by considering open covers where each element of the cover represents an interval on which it is "easy" to track the roots, since no crossing can occur.

For any $\xi$ define the number $f(\xi)$ to equal the minimum distance between a pair of distinct roots $\lambda_1$ and $\lambda_2$ of $p(\lambda,\xi)$. This is always a positive finite number. Now, choose some $\varepsilon(\xi) = \frac{f(\xi)}6$ and find some corresponding $\delta(\xi)$ such that the roots of the polynomials $p(\lambda,\xi')$ for $|\xi'-\xi|<\delta(\xi)$ always lie strictly within $\varepsilon(\xi)$ of those of $p(\lambda,\xi)$ in the sense of the theorem.

Then, the trick is to choose some sequence $\xi_1<\xi_2<\xi_3<\ldots<\xi_n$ such that $(\xi_i-\delta(\xi_i),\xi_i+\delta(\xi_i))$ is a cover of $[a,b]$. Then, we prove some lemmas, where we take $\xi$ and $\xi'$ to be such that $|\xi-\xi'| < \delta(\xi)+\delta(\xi')$ as is the case for consecutive terms of the above sequence.

If $\lambda_1$ and $\lambda_2$ are distinct roots of $p(\lambda,\xi)$ and $\lambda_1'$ and $\lambda_2'$ are distinct roots of $p(\lambda,\xi')$ such that $|\lambda_1-\lambda_1'| < \varepsilon(\xi)+\varepsilon(\xi')$ and $|\lambda_2-\lambda_2'| < \varepsilon(\xi)+\varepsilon(\xi')$, then $|\lambda_1-\lambda_2'| > \varepsilon(\xi)+\varepsilon(\xi')$.

This is mostly tedious. It's also where the coefficient of $\frac{1}6$ comes into $\varepsilon(\xi)$. Basically, this says that if we consider a graph consisting of an edge between every pair $(\lambda,\lambda')$ where $p(\lambda,\xi)=p(\lambda',\xi')=0$ and $|\lambda-\lambda'| < \varepsilon(\xi)+\varepsilon(\xi')$, then every connected component is either a single root $\lambda$ of $p(\lambda,\xi)$ connected to some number of roots of $p(\lambda',\xi')$ or is a single root of $p(\lambda',\xi')$ connected to some number of roots of $p(\lambda,\xi)$. This means that, however we labelled the roots of $p(\lambda,\xi)$, there is a more or less unique way to label the roots of $p(\lambda,\xi')$.
Basically, the rest of the proof from here is to choose some new sequence $\xi'_1<\xi'_2<\xi'_3<\ldots<\xi'_{n'}$ which includes the old one and extend the labeling in some way consistent with the old one and to continually do this. We can use this to construct a labeling on some dense set of the $\xi$ and then extend continuous to get continuous functions tracing the roots of the family of polynomials.
There's a lot of places to fill in more analytical rigor to this argument, but hopefully this sketch is helpful to you. I can flesh out any parts that are unclear.
