We first start with following definitions.
Definition 1. A family $\mathcal{P}$ of seminorms on a real vector space $X$ is called filtering if for any $p_1,p_2\in \mathcal{P}$ there exsist $q\in \mathcal{P}$ and $r_1,r_2>0$ such that the two inequalities $r_1p_1\le q$ and $r_2p_2\le q$ hold on $X$.
Definition 2. A family $\mathcal{P}$ of seminorms on a real vector space $X$ is called separating if for every $x\in X$ with $x\neq 0$, there exists $p\in \mathcal{P}$ such that $p(x)>0$.
My question is this: Suppose that $(X,\tau)$ is a Hausdorff locally convex space. How can we construct a family of seminorms on $X$ which is both separating and filtering?
My idea. If $(X,\tau)$ is a Hausdorff locally convex space, then as pointed out in Rudin's Functional Analysis book, we can construct a separating family $\mathcal{P}$ of seminorms on $X$. At this point, I don't have any idea if $\mathcal{P}$ is filtering.
Tips are very much appreciated.
Since you only want a filtering family of seminorms you could simply take the family of all continuous seminorms $$ \mathcal{P} = \{p \colon X \to [0,\infty) \mid p \text{ is continuous}\} $$ this is clearly a filtering family and it is separating if and only if $X$ is Hausdorff.
More useful is the following construction: Let $\mathscr{U}$ a neighborhood basis of $0$ consisting of open, convex, balanced, circled and absorbing sets and for $U \in \mathscr{U}$ let $p_U(x) = \inf\{\lambda \geq 0 \mid x \in \lambda U\}$ be the Minkowski functional associated with $U$. Then $$ \mathcal{P} = \{p_U \mid U \in \mathscr{U}\} $$ is a family of continuous seminorms on $X$.
The family induces the given locally convex topology $\tau$ on $X$: Since all $p_U$ are continuous, the topology induced by $\mathcal{P}$ is weaker than $\tau$ and since $U = \{x \in X \mid p_U(x) \lt 1\}$, the topology induced by $\mathcal{P}$ is at least as strong as $\tau$.
If $X$ is Hausdorff, $\mathcal{P}$ is separating: if $x \neq 0$ there is $U \in \mathscr{U}$ such that $x \notin U$, so $p_U(x) \geq 1$.
The family $\mathcal{P}$ is filtering since for $U_1, U_2 \in \mathscr{U}$ there is $V \in \mathscr{U}$ such that $V \subseteq U_1 \cap U_2$ by the definition of a neighborhood basis, and therefore $p_V \geq p_{U_1}, p_{U_2}$.