Let us consider the space $C^0([0,1])$ of real continuous functions over $[0,1]$.
Let $A = \{a_n : n \in \mathbb{N}\}$ be a countable set of $[0,1]$ and $\alpha_n$ be strictly positive real numbers such that $\sum_{n=0}^{+\infty} \alpha_n < +\infty.$ Now we define a seminorm by $$||f||_{A,\alpha} = \sum_{n=0}^{+\infty} \alpha_n |f(a_n)|.$$
If we have two such seminorms $||.||_{A,\alpha}$ and $||.||_{A',\alpha'}$ which are equivalent, I would like to show that necessarily $A=A'.$ I managed to obtain $\overline{A} = \overline{A'}$ but I have difficulties to obtain the desired equality.
We argue by contradiction. Assume that $a_n \notin A'$ for some $n$. Let $f_{n, \delta} = \max \{ 0, 1- \delta^{-1}|x-a_n|\}$ (It's $1$ at $a_n$ and is zero outside $[a_n - \delta, a_n+\delta]$). Note that
$$\| f_{n, \delta}\|_{A, \alpha}\ge \alpha_n.$$
For each $\epsilon >0$, let $M_\epsilon$ be large so that $$\sum_{M_{\epsilon}}^\infty \alpha_n' <\epsilon.$$
Then find $\delta$ small enough so that $|a_m' - a_n| > \delta$ for all $m=1, \cdots, M_\epsilon$ (here we used that $a_n \neq A'$). Then $$ \|f\|_{A', \alpha'} = \sum_{M_\epsilon}^\infty |f(a_n')| \le \epsilon.$$
Thus for all $\epsilon>0$ we have found continuous functions so that $\| f_{n, \delta}\|_{A, \alpha}\ge \alpha_n$ and $\| f_{n, \delta}\|_{A', \alpha'}\le \epsilon$. Thus the two seminorms cannot be equivalent.