I am trying to show that $\mathbb{R}$ is a first countable space with respect to the particular point topology defined by:
$$\mathcal{T} = \{I\subseteq \mathbb{R} : I = \emptyset \text{ or } p\in I\}$$
for an arbitrary point $p\in \mathbb{R}$. My approach was to find a countable neighbourhood basis of a point $x\in \mathbb{R}$ other than $p$. I believe that the subset consisting of the closed interval $\mathcal{B}_x := \{[x,p]\}$ does the job, as any neighbourhood of $x$ must contain the closed interval (assuming that $x<p$, otherwise we just take the closed interval $[p,x]$ as the basis element instead) $[x,p]$, and the closed interval is considered open under the particular point topology. Clearly $\mathcal{B}_x$ is countable as its cardinality is $1$, and so it is a countable neighbourhood basis, which implies first countability.
Question: Is my proof valid? I was checking my proof with chatgpt but it keeps telling me that I should have a collection rather than a single set, and it also said that the set I provided is not open, which is clearly false by the definition of the topology above.
In this topology $\{x,p\}$ is an open set containing $x$. Instead of $\mathcal B_x=[x,p]$ or $\mathcal B_x=[p,x]$ you have to take $\mathcal B_x=\{\{x,p\}\}$.