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Prove $X$ is closed in $Y \iff \mathcal{F}_X$ is closed in $Y$.

Can anybody help me on this proof of the following proposition:

$X$ is closed in $Y$ if and only if $\mathcal{F}_X$ is closed in $Y$, where $\mathcal{F}_X$ is the filter generated by $X$.

I’ve got this:
$X$ is closed in $Y$ iff there is a collection $\mathcal{C}$ that is a neighborhood base for $Y$ and such that $X \subseteq \mathcal{C}$, namely $X$ is the unique intersection of $\mathcal{C}$.
Now $\mathcal{F}_X$ is closed in $Y$ iff there is a collection $\mathcal{

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