Let $X$ be a vector space. For each $x \in X$, let $\theta_x \colon X \rightarrow X$, $\theta_x(y) = x + y$. Consider $\theta = \{\theta_x\}_{x \in X}$. Now, consider $p = \{p_{w}\}_{w\in O}$ a family of seminorms in X that separates points (that is, if $x \neq 0$, there is $w \in O$ such that $p_{w}(x) > 0$).
I have to show that $(X, \sigma(X, p \circ \theta))$ is a Hausdorff space, where $S_p(0) = \{p_w^{-1}([0, n^{-1})) : w \in O, n \in \mathbb{N}\}$ is a sub-base of neighborhoods of $0$ in $\sigma(X, p\circ\theta)$.
Consider $x, x' \in X$, $x \neq x'$. I need to show that there are $A_x$, $A_{x'}$ open sets in $\sigma(X, p\circ\theta)$ such that $x \in A_x, x' \in A_{x'}$ and $A_x \cap A_{x'} = \emptyset$.
As $p$ separates points and $x - x' \neq 0$, there is $w \in O$ such that $p_{w}(x - x') > 0$ and we have that
$$0 < p_w(x - x') \leq p_w(x) + p_w(-x') = p_w(x) + p_w(x').$$
Let $r > 0$ such that $p_w(x) + p_w(x') = r$ and let $n \in \mathbb{N}$ be such that $n^{-1} < \frac{r}{2}$.
Claim. $A_x = \theta_x \circ p_w^{-1}([0, n^{-1}))$ and $A_{x'} = \theta_{x'} \circ p_w^{-1}([0, n^{-1}))$ satisfy the necessary conditions.
Clearly, $x \in A_{x}$, $x' \in A_{x'}$ and $A_x$ and $A_{x'}$ are open sets (see the definition of S_p and notice that for all $y \in X$, $\theta_y$ is an homeomorphism). It only remains to show that $A_x \cap A_{x'} = \emptyset$.
I tried to show this by contradiction, but I am having some problems...
I suspect you are confusing yourself a bit with too much notation. This isn't that different from checking that the topology induced by a norm is Hausdorff.
Since translations are homeomorphisms you might as well assume for simplicity that $x' = 0$. There is no loss of generality here because if $A$ is an open neighbourhood of $0$ and $B$ is an open neighbourhood of $x-x'$ such that $A \cap B = \emptyset$ then $\theta_{x'}(A)$ and $\theta_{x'}(B)$ are disjoint open neighbourhoods of $x'$ and $x$ respectively.
So suppose $x \neq 0$. Then there is a $w \in O$ such that $p_w(x) > 0$. From here we can proceed exactly as in the case that $p_w$ is a norm and the topology on $X$ is the norm topology.
Pick $n$ such that $n^{-1} < p_w(x)$. I claim that if $A = \{y : p_w(y) < (2n)^{-1}\} = p_w^{-1}[0,\frac1{2n})$ then $A$ and $\theta_x(A)$ are disjoint. If not, there is $y \in A \cap \theta_x(A)$ which means that $p_w(y) < \frac1{2n}$ and $p_w(y-x) < \frac1{2n}$. Then, by the triangle inequality, $p_w(x) \leq p_w(y) + p_w(x-y) < \frac{1}{n}$ which is a contradiction.