The space of continuous functions on an interval has a countable dense subset and a countable basis Give $\Bbb R^I$ the uniform metric, where $I = [0, 1]$. Let $C(I, \Bbb R)$ be the subspace consisting of continuous functions. Show that $C(I, \Bbb R)$ has a countable dense subset, and therefore a countable basis. (Source: Munkres Question 15 Article -30)
My try:
Consider the set of all polynomials $\mathscr{P}=\{a_0+a_1x+...+a_nx^n:a_i\in \Bbb Q;n\in \Bbb N\}$ on $[0,1]$. By the Weierstrass Approximation Theorem, for every continuous function $f$ there exists a sequence of polynomials $p_n \rightarrow f$ (uniformly).Hence $\scr P$ is dense in $C(I,\Bbb R)$.
Also $C(I, \Bbb R)$ is metrizable and hence a countable dense subset and hence has a countable basis.
Is my proof correct?

Yes, if the Weierstrass theorem has been covered, this does follow. 
