Let $X$ be a locally convex space, whose topology is defined by a family of seminorms $\mathcal{P}$.
Fact. For every continuous seminorm $q: X\to \mathbb{F}$ and positive number $\epsilon>0$, the neighborhood of the origin
$V_{q, \epsilon}:=\{x\in X: q(x)< \epsilon\}$ contains a basic open set of the following form $$U_{p_1, \cdots, p_N; \eta}:=\{x\in X: \max\{p_1(x), \cdots, p_N(x)\}< \eta, \eta>0, p_j\in \mathcal{P}\}.$$
Question. Can we conclude, if $x\in X$ satisfies $p_1(x)=\cdots=p_N(x)=0$, then $q(x)=0$? (The point is, if we let $\epsilon\to 0^+$, the seminorms $p_1, \cdots, p_N$ are fixed while only $\eta\to 0^+$.)
Attempted proof. Let $\theta>0$ and $x'\in U_{p_1, \cdots, p_N; \theta\eta}$. Then, by the property of seminorms, we have $$\max\left\{p_1\left(\frac{x'}{\theta}\right), \cdots, p_N\left(\frac{x'}{\theta}\right)\right\}<\eta\Leftrightarrow \frac{x'}{\theta}\in U_{p_1, \cdots, p_N; \eta}\subset V_{q,\epsilon} \Rightarrow x'\in V_{q, \theta\epsilon}.$$ That is, $$U_{p_1, \cdots, p_N; \theta\eta}\subset V_{q, \theta\epsilon}.$$ Hence, if $x\in X$ satisfies $p_1(x)=\cdots=p_N(x)=0$, it also satisfies $$x\in \bigcap_{\theta>0}V_{q, \theta\epsilon}=\text{Ker}(q).$$ That is, $q(x)=0$. $\square$
Is the above reasoning correct? Thanks in advance.
Background: I'm trying to understand the proof of the following proposition.
For a normed space $X$, the weak*-topology of the dual $X^*$ is metrizable iff $X$ has a finite or countable Hamel basis.
One step in the proof claims the conclusion of my question, where the seminorm is the canonical pairing $p_x(\cdot):=|\langle \cdot, x\rangle|$. See here and here