I wish to prove that if a functional $\tau$ on some vector space $X$ is continuous in a locally convex topology (meaning one generated by some family of seminorms $\Gamma$) then there is a constant $M$, (possibly dependent on $n$) we have for any $x \in X$ and some $n\in\mathbb{N}$)
$$|\tau(x)|<M \max_{1<j<n} p_j(x)$$
The proof (Murphy page 267) goes as follows The set $S:=\{x\in X \ |\tau(x)|<1\}$ is a neighborhood of $0$ in $X$. This is pretty obvious. Then we say there are semi-norms $p_1...p_n$ and $\epsilon>0$ such that $U(\epsilon, p_1...p_n)\subseteq S$ where the $U(\epsilon, p_1...p_n)=\{u\in X| p_j(u)<\epsilon \ \ j\in \{1,...n\} \}$.
But then he says that for $p=\max_{1<j<n} p_j$ if $p(x)<\epsilon$ then it implies $|\tau(x)|<1$. I'm confused about what this means, surely not that the fact that all of the seminorms in this neighbourhood that we defined post factum somehow implies that the norm is indeed less than one (since it is simply a thing we chose to begin with arbitrarily).
What is the meaning of this implication?
The way you are writing things might be confusing you.
The situation is that you have a continuous linear functional $\tau$. The claim is that then one can find $M>0$ and seminorms $p_1,\ldots,p_n$ such that $$|\tau(x)|\leq M\,\max\{p_1(x),\ldots,p_n(x)\}. $$
For this one defines the set $S$ as you mention, which is open by hypothesis since $\tau$ is continuous. This means, as you say, that you can find a basic neighbourhood $U(\epsilon; p_1,\ldots,p_n)\subset S$. He puts $$ p(x)=\max\{p_1(x),\ldots,p_n(x)\}. $$ If $p(x)<\epsilon$, this means that $p_j(x)<\epsilon$ for all $j$, and thus $x\in U(\epsilon;p_1,\ldots,p_n)$, which is a subset of $S$. So we have that $x\in S$ and this means that $|\tau(x)|<1$.