Group acting on a topology

Group acting on a topology

Let $G$ be a group acting on a set $X$ under a topology $T$. Assume that
the action is faithful, that is, no $1\ne g\in G$ fixes every $x\in X$.
Denoting $\{y^g:y\in Y\}$ by $Y^g$, we say that $T$ is $g$-fresh if
$\{Y^g:Y\in T\}=: T^g$ is a topology on $X$. If $T$ is $g$-fresh for every
$g\in G$, then $T$ is $G$-fresh.
It's not hard to see that $X$ and $\emptyset$ are always $g$-fresh.
Additionally,
$$\begin{align*} \left(\bigcup_{\alpha\in A}u_\alpha\right)^g&= \{x^g:x\in
u_\alpha \text{ for some } \alpha\in A\}\\ &=\{x:x\in (u_\alpha)^g \text{
for some } \alpha \}\\ &=\bigcup_{\alpha\in A} \left(u_\alpha\right)^g
\end{align*}$$
so unions are sent to unions under the action of $g$. If we can show that
finite intersections in $T^g$ are contained in $T^g$, we'll have that the
image of every topology under the action of a group on its underlying set
is a topology. Is this true? If not, please provide a counterexample. (is
there one that is minimal in some way?)