Denote $\mathbb{F}[x]$ (where $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$) is the space consisted of polynomials with coefficients in the field $\mathbb{F}$. The question is whether there is a metric $d$ such that $(\mathbb{F}[x], d)$ is completed.
My first thought to this problem is no. If there do exists such metric, then $(\mathbb{F}[x], d)$ must be in the second Baire’s category. Meanwhile $\mathbb{F}[x]$ is the countable infinite union of $\mathbb{F}^n[x]$, the set of all polynomials with coefficients in the field $\mathbb{F}$ and has degree no greater than $n$, and since $\mathbb{F}^n[x]$ is nowhere dense (which I can not prove), $(\mathbb{F}[x], d)$ is in the first Baire’s category, a contradiction.
The problem now is to prove that $\mathbb{F}^n[x]$ is nowhere dense in $\mathbb {F}[x]$. My idea is to show that if $\mathcal{O}$ is an open subset of $\mathbb {F}[x]$, then it must contains some polynomial of arbitrary degree. Hence $\mathbb{F}^n[x]$ is nowhere dense in any open subset $\mathcal{O}$. But the thing the topology is unsure since we don’t know $d$, and I’m stuck from here.
Is my intuition correct? If it do, how should I proceed from here?
Yeah, you cannot prove that because that is not necessarily true.
Unfortunately it is not. Without additional assumptions on the metric $\mathbb{F}[x]$ is just an uncountable set. In no way different from $\mathbb{F}$. In particular there are many, many different complete metrics on it. You can put a discrete metric on it, you can put a metric by transfering it from $\mathbb{F}$ or from an uncountable compact space if you wish. You do that by considering a bijection $f:\mathbb{F}[x]\to X$ and defining $d_{\mathbb{F}[x]}(P,Q)=d_X(f(P),f(Q))$.
The fact that $\mathbb{F}[x]$ is a ring of polynomials does not matter at all. You did not impose any condition that would relate this algebraic structure to the metric structure. And thus the number of solutions is infinite.