By the Weierstrass approximation theorem for $f\in C[a,b]$ there exists a sequence ($Q_n$) of polynomials such that $Q_n(x) \rightrightarrows f(x)$ on $[a,b]$ or equivalently for each $\varepsilon >0$ there exists a polynomial $Q$ such that $|Q(x)-f(x)|<\varepsilon$.
Let $P_n$ be athe space of all polynomials (one real variable) of degree $\leq n$, and $E_n(f)=dist(f,P_n)$.
How to show using the Weierstrass theorem that for continuous $f: [a,b]\rightarrow \mathbb R$ : $$lim_{n\rightarrow \infty} E_n(f)=0.$$
The Weierstrass theorem says nothing about degrees of approximating $f$ polynomials.
It's obvious that $\bigl(E_{n}(f)\bigr)$ is non-increasing in $n$, and Weierstrass approximation says $E_{n}(f)$ can be made arbitrarily small by taking $n$ sufficiently large.