First I'd like to say that although this question was asked before (here) and is from the same text, the answer used methods that were not introduced in the text.
Let $P_n(x)$ be a polynomial of degree $n$ and $f:[a,b] \to \Bbb R$ a bounded function.
Define $$\Delta(P_n) = \sup_{x\in[a,b]}|f(x) -P_n(x)| \ \ \mathrm{and} \ \ E_n(f) =\inf_{P_n} \Delta(P_n),$$ the latter being taken over all polynomials of degree $n$. A polynomial is called a polynomial of best approximation of $f$ if $\Delta(P_n) = E_n(f)$
The previous questions are:
$a)$ show a polynomial of best approximation of degree $0$ exists;
$b)$ among the polynomials $Q_{\lambda}(x)$ of the form $\lambda P_n(x)$, where $P_n$ is a fixed polynomial, there is a polynomial $Q_{\lambda_0}$ such that $$\Delta(Q_{\lambda_0}) = \min_{\lambda \in \Bbb R}\Delta(Q_{\lambda}).$$
The question I am on is:
$c)$ if there exists a polynomial of approximation of degree $n$, there also exists a polynomial of best approximation of degree $n + 1$.
This last part looks like an inductive step which would use $b)$ with $f$ as $f(x) - P_n(x)$ and $Q_{\lambda}$ as $\lambda x^{n+1},$ $P_n(x)$ being the best approximation. In general, though, the polynomial of degree $n+1$ is not in the form of $P_n(x) + \lambda_0 x^{n+1}.$ Because of this I'm not sure how the question can be done.