Answer by bof for Conjecture about union-closed families of sets - attempt 2

To get a counterexample to the current version of your conjecture, set $t=2$ and $s=6$ in the following generalization of my counterexample to your previous question.

Theorem. Given $t\in\mathbb N$ and $\varepsilon\gt0$, we can construct a nonempty finite union-closed family $\mathcal F$ of nonempty sets with a subfamily $\mathcal H\subseteq\mathcal F$ satisfying the conditions:
(i) $H\in\mathcal H\implies\frac{|\mathcal P(H)\cap\mathcal F|}{|\mathcal F|}\gt\frac23-\varepsilon$;
(ii) $F\in\mathcal F\setminus\mathcal H\implies\frac{|\mathcal P(F)\cap\mathcal F|}{|\mathcal F|}\lt\frac13$;
(iii) $F\in\mathcal F\implies|\{H\in\mathcal H:F\not\subseteq H\}|\ge t$.

Proof. Choose $s\in\mathbb N$ so that $\frac{2s+1}{3s+3t+1}\gt\frac23-\varepsilon$.

Choose disjoint sets $A,B,C,X,Y,Z$ with $|A|=|B|=|C|=s$ and $|X|=|Y|=|Z|=t$; let $A=\{a_1,\dots,a_s\}$, $B=\{b_1,\dots,b_s\}$, $C=\{c_1,\dots,c_s\}$, $X=\{x_1,\dots,x_t\}$, $Y=\{y_1,\dots,y_t\}$, $Z=\{z_1,\dots,z_t\}$.

For $i=0,1,\dots,s-1$ define:
$A_i=X\cup B\cup C\cup\{a_j:j\le i\}$;
$B_i=Y\cup A\cup C\cup\{b_j:j\le i\}$;
$C_i=Z\cup A\cup B\cup\{c_j:j\le i\}$.

For $i=0,1,\dots,t-1$ define:
$X_i=Y\cup Z\cup A\cup B\cup C\cup\{x_j:j\le i\}$;
$Y_i=X\cup Z\cup A\cup B\cup C\cup\{y_j:j\le i\}$;
$Z_i=X\cup Y\cup A\cup B\cup C\cup\{z_j:j\le i\}$.

Finally, define
$T=X\cup Y\cup Z\cup A\cup B\cup C$
and let
$\mathcal F=\{A_j:0\le j\lt s\}\cup\{B_j:0\le j\lt s\}\cup\{C_j:0\le j\lt s\}\cup\{X_j:0\le j\lt t\}\cup\{Y_j:0\le j\lt t\}\cup\{Z_j:0\le j\lt t\}\cup\{T\}$.

Then $\mathcal F$ is a union-closed family with $|\mathcal F|=3s+3t+1$. Let
$\mathcal H=\{X_j:0\le j\lt t\}\cup\{Y_j:0\le j\lt t\}\cup\{Z_j:0\le j\lt t\}\cup\{T\}\subset\mathcal F.$

(i) $H\in H\implies\frac{|\mathcal P(H)\cap\mathcal F|}{|\mathcal F|}\ge\frac{2s+1}{3s+3t+1}\gt\frac23-\varepsilon$;
(ii) $F\in\mathcal F\setminus\mathcal H\implies\frac{|\mathcal P(F)\cap\mathcal F|}{|\mathcal F|}\le\frac s{3s+3t+1}\lt\frac13$;
(iii) $\{H\in\mathcal H:A_0\not\subseteq H\}\supseteq\{X_0,\dots,X_{t-1}\}$,
$\{H\in\mathcal H:B_0\not\subseteq H\}\supseteq\{Y_0,\dots,Y_{t-1}\}$,
$\{H\in\mathcal H:C_0\not\subseteq H\}\supseteq\{Z_0,\dots,Z_{t-1}\}$.