Here’s the statement I want to prove:
If $P$ is a prime ideal in a commutative ring $R$ with unity, and $\{P\}$ is closed under the Zariski topology on $\text{Spec}R$, then $P$ is maximal.
Here, for an ideal $J$ of $R$, we define a closed subset to be $$V(J):= \{P \in \text{Spec}R \ | \ P \supseteq J\}$$
I want to ask for opinions on my attempt below.
Assume for a contradiction that $P$ is not maximal. So $P$ can be enlarged, i.e., there exists $k \in R \setminus P$ such that $(k) + P \neq R$. In particular, $(k) + P$ is a proper ideal that is strictly larger than $P$. Let $J \subseteq P$, where $J$ is an ideal of $R$. Then $V(J) \ni(k) + P$, which implies there is no ideals $J$ of $R$ such that $V(J) = \{P\}$. Hence $\{P\}$ is not closed. $\Rightarrow$ Contradiction.