Prove that if $\mathcal{F}[x_d]\subseteq \mathcal{G}$, then there's a subnet $(y_e)$ of $(x_d)$ such that $\mathcal{G}=\mathcal{F}[y_e]$.

Question

Let $X$ be a set.

Given any net $(x_d)_{d\in D}$ in $X$, we define

$$\color{red}{\mathcal{F}[x_d]}:=\big\{F\in 2^X:(\exists d_0\in D)(\forall d\in D)\big(d_0\preceq _Dd\Rightarrow x_d\in F\big)\big\}.$$

It's easy to prove that $\mathcal{F}[x_d]$ is a filter on $X$.

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My question is: Prove that if $\mathcal{G}$ is a filter on $X$ such that $\mathcal{F}[x_d]\subseteq \mathcal{G}$, then there's a subnet $(y_e)_{e\in E}$ of $(x_d)_{d\in D}$ such that $\mathcal{G}=\mathcal{F}[y_e]$.

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I'm using the following definition of subnet.

We say that $(y_e)_{e\in E}$ is a subnet of $(x_d)_{d\in D}$ is there's a map $\kappa :E\to D$ such that

1. $y_e=x_{\kappa(e)}$ for all $e\in E$
2. $\kappa (e_1)\preceq_D\kappa(e_2)$ for any $e_1,e_2\in E$ with $e_1\preceq _Ee_2$;
3. For all $d\in D$ there's $e\in E$ such that $d\preceq _D\kappa (e)$

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I tried a standard approach to construct a subnet of $(x_d)_{d\in D}$. That is, first I tried to find a proposition $\phi $ such that $E:=\{(d,F)\in D\times \mathcal{G}:\phi \}$ is a directed set with the relation $\preceq_E $ given by: $(d_1,F_1)\preceq _E(d_2,F_2)$ if and only if $d_1\preceq _Dd_2$ and $F_2\subseteq F_1$. Then define $\kappa :E\to D$ by $\kappa (d,F):=d$ and $y:E\to X,\,e\mapsto y_e$ by $y_e:=x_{\kappa(e)}$.

However I didn't find a convenient proposition $\phi$ such that $\mathcal{G}=\mathcal{F}[y_e]$.

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Thank you for your attention