There exist a degree $n+1$ polynomial of best approximation if there exist a degree $n$ polynomial of best approximation This is a problem from Mathematical Analysis by Zorich, from the chapter on continuous functions.

Definition:
Let $P_n$ be a polynomial of degree $n$. For a function $f:[a,b]\to\mathbb{R}$, Let $\Delta(P_n) = \sup_{x\in[a,b]} |f(x)-P_n(x)|$. and $E_n(f) = \inf_{P_n} \Delta(P_n)$. A polynomial $P_n$ is the best approximation of degree $n$ of $f$ is $\Delta(P_n) = E_n(f)$.

I have already proved the following:

There exist a polynomial $P_0(x) = a$ of best approximation of degree 0.

If $Q_\lambda(x) = \lambda P_n(x)$ for some fixed polynomial $P_n$. Then there exist a polynomial $Q_{\lambda_0}$ such that $\Delta(Q_{\lambda_0}) = \min_{\lambda\in \mathbb{R}} \Delta(Q_\lambda)$

I'm stuck on proving the following:

If there exists a polynomial of best approximation of degree $n$, there also exists a polynomial of best approximation of degree $n+1$.

My intuition is to prove $\lambda_0 x^{n+1}+P_n$ is the $n+1$ best approximation for $f$. Where $\Delta(\lambda_0 x^{n+1}) =  \min_{\lambda\in \mathbb{R}} \Delta(\lambda x^{n+1})$ and $\Delta(P_n) = E(f(x) - \lambda_0 x^{n+1})$.

I don't know if this approach is right. I'm stuck on how to proceed.
Determining the best $n$th degree approximation explicitely is a very difficult problem and cannot be done by a simple induction (see the chapter on Tschebyscheff approximation in your favorite textbook of numerical analysis). But existence is easy, it follows from general principles: You have to set up a certain continuous function $\Phi$ on a certain compact set $K$ in a high-dimensional space.
I also do not see how to prove the induction step, but it seems easier instead to prove the existence of polynomial of best approximation for all natural degrees. 
One of possible ways to follow those principles, mentioned by  @ChristianBlatter is below.



$\alpha, \beta \in \mathbb{R}^n$ 

$P(\alpha) = \alpha_0x_0 + ...+\alpha_nx_n$  

We can introduce a metric $d$ on bounded functions at $[a,b]$: 
$$d(f,g) = \sup_{b \le x \le a} |f(x) - g(x)|$$

Denote $\delta \in \mathbb{R}^{&gt;0}$ - neighborhood of point $\alpha \in \mathbb{R}^n $  related to metric $d$ as $U^{\delta}_{d}{(\alpha)}$

Let's consider triangle inequality for $d$ :
$$d(f(x), P(\alpha)) \le d(f(x), P(\beta)) + d(P(\alpha), P(\beta) ) \tag{1} $$ 
$$ \Rightarrow |d(f(x), P(\alpha)) - d(f(x), P(\beta))| \le d(P(\alpha), P(\beta) ) \tag{2}$$

Assume that we have chosen some fixed $\alpha$. Left-hand side of last inequality is actually bounded from the top by oscillation of function
 $\Delta(\alpha) :=  |\sup_{x \in [a,b]} {(f(x) - P(\alpha))} |$ at $d(\alpha, \beta)$ - neighbourhood of point $\alpha$ or shortly: 
$\le \omega_{\Delta(x)}( U^{d(\alpha, \beta)}_{d}{(\alpha)} )$. In other words, given $P(\beta)$ close enough to $P(\alpha)$, their images under $\Delta$ can become as close as we want. 

That is why $\Delta$ is continuous map from $\mathbb{R}^n$ to $\mathbb{R}$.
Continious function on set is continuous on subset, so our next goal is to construct a compact subset of $S \subset \mathbb{R}^n$ (each point representing a polynomial of degree $\le n$) such that 
$\forall x \in {\mathbb{R}^n}\setminus \{S\}$ and $ \exists  y \in S$ that: $ \Delta(y) \le \Delta(x)$ (this condition means that minimum of $\Delta$ is certainly NOT in ${\mathbb{R}^n}\setminus \{S\}$ if exists). 

Let $S$ be a zero-centered ball (with border) in ${\mathbb{R}^n}$ bounded with sphere of radius $r_0 &gt; 2|\sup_{x \in [a,b]} {f(x)}|$ in metric space with metric $d$.
Obviously, $\forall x  \in {\mathbb{R}^n}\setminus S : d(f,x) \ge |\sup_{x \in [a,b]} {f(x)}| = d(0, f)$ 
In other words, $0 \in {\mathbb{R}^n}$ is closer to $f$ than any polynomial outside $S$. $\implies$ If minimum of exists, it is in $S$!

$S$ is compact, $\Delta$ is continuous $\implies$ minimum exists.

q.e.d