First countable property for $\operatorname{spec}(\mathbb{C}[x])$ with Zariski topology I'm interested in this particular property which i've developed a proof and i would like to know if my thoughts are correct.
I've only considered the subspace $X = \operatorname{spec}(\mathbb{C}[x])\setminus \{(0)\}$, and since $X$ is the set of the maximal ideals we obtain it has the cofinite topology (as subspace topology). Using together $X$ is non countable then it doesn't have countable local basis. Hence $\operatorname{spec}(\mathbb{C}[x])$ is not first countable, because if it was $X$ would be too.
I think that the major part of my doubt is in using that $X$ has the cofinite topology as subspace topology. Obtain that the cofinite topology is a subset of the subspace topology it's ok, but the other inclusion makes me feel unconfortable and i don't know how to do it properly. Any help will be extremely appreciated.

Edit: An attempt of a more solid answer for the equality between those topologies. It's clearer that the cofinite topology is contained in the zariski space topology (z.s.p.) since any maximal ideal is a closed point. If we take an open set $A$ of the z.s.p. then exists an open set $A'$ in the zariski topology that $A = A' \cap X$. We have that $\mathbb{C}[x]$ is a P.I.D. then $(A')^c$ as closed set must be given by a principal ideal, with generator being a polynomial of $n$ degree. Using that the complex numbers are algebraically closed then we can decompose it into  $n$ linear factors which will give us our closed set $(A')^c$ in terms of a finite union of maximal ideals (as closed points), then granting that $A$ is the complement of a finite set, hence it must be an open set of the cofinite topology.
Is this proof correct? I can write down with more symbols and details if it is incomprehensible.
 A: If indeed we have a set $X$ in the cofinite topology, then $X$ is first countable iff $X$ is countable.
The right to left implication is trivial: if $X$ is countable, $X$ has only countably many open sets (because there are only countably many finite sets, plus $X$ so countably many closed sets as well), so is trivially first and second countable.
And if $X$ is uncountable, $p \in X$, and $\mathcal{B}$ is a countable family of open sets all containing $p$, then all $X\setminus B$, $B \in \mathcal{B}$ are finite, and so
$$N= \bigcup \{X\setminus B: B \in \mathcal{B}\} \cup \{p\}$$
is countable (a countable union of finite sets) and so there is some $q \in X\setminus N$, as $X$ is uncountable. But then $X\setminus \{q\}$ is open, contains $p$, but does not contain any $B \in \mathcal{B}$, because $B \subseteq X\setminus \{x\}$ implies $x \in X\setminus B \subseteq N$, which would be a contradiction (as $x \notin N$). So $\mathcal{B}$ is not a local base at $p$. This shows that $X$ is not first countable at any point of $X$.
