Question
I came across the above claim in a proof of Lyons, Lévy, and Caruana, here it is largely verbatim:
"Indeed, if two elements $\mathbf{s}_{\mathbf{1}}$ and $\mathbf{s}_{\mathbf{2}}$ of $T((E))$ are distinct, there exists [a smallest] $n \geq 0$ such that $\pi_n\left(\mathbf{s}_{\mathbf{1}}\right) \neq \pi_n\left(\mathbf{s}_{\mathbf{2}}\right)$. Since $T^{(n)}(E)$ is finite dimensional, there exists a linear form $\sigma \in T^{(n)}(E)^*$ which separates $\pi_n\left(\mathbf{s}_{\mathbf{1}}\right)$ and $\pi_n\left(\mathbf{s}_{\mathbf{1}}\right)$".
Here, $T((E)):= \bigoplus_{k\ge0}E^{\otimes k},$ where $E^{\otimes0}:=\mathbb R$, while $T^{(n)}(E) = \pi_n(\,T((E))\,)$ is the $n$-th level truncation. So the linear form can be just taken to be the $n$-th level projection, composed with a linear form $e^*:V^{\otimes n}\to \mathbb R$ that picks up the coefficient in the basis of $V^{\otimes n}$ where $\text{proj}_n(s_1)$ and $\text{proj}_n(s_2)$ differ.
My question is, why do we need to assume that the dimension of the vector space is finite to find a linear form that separates points? I would hazard the following guess: when we consider infinite dimensional spaces, assuming two elements are just different doesn't tell us how to construct a form because, since there may be no known basis for the space (?without the axiom of choice?), there is no way of comparing the two elements and finding out "where" (in the basis decomposition) they differ, allowing us to construct a linear form that "picks up" this difference.
Background
Full context I drew this claim from:
Corollary 2.16. If the signatures of a finite collection of paths with finite $p$-variation for some $p<2$ are pairwise distinct, then they are linearly independent. Proof. Let $X_1, \ldots, X_n$ be $n$ paths with bounded variation and pairwise distinct signatures. On the finite set $\left\{S\left(X_1\right), \ldots, S\left(X_n\right)\right\}$, the elements of $T\left(E^*\right)$ induce an algebra of real-valued functions. This algebra contains the constants, because the first term of every signature is 1. Moreover, it separates the points. Indeed, if two elements $\mathbf{s}_{\mathbf{1}}$ and $\mathbf{s}_{\mathbf{2}}$ of $T((E))$ are distinct, there exists $n \geq 0$ such that $\pi_n\left(\mathbf{s}_{\mathbf{1}}\right) \neq \pi_n\left(\mathbf{s}_{\mathbf{2}}\right)$. Since $T^{(n)}(E)$ is finite dimensional, there exists a linear form $\sigma \in T^{(n)}(E)^*$ which separates $\pi_n\left(\mathbf{s}_{\mathbf{1}}\right)$ and $\pi_n\left(\mathbf{s}_{\mathbf{1}}\right)$. Now, $T^{(n)}(E)^*=T^{(n)}\left(E^*\right) \subset T\left(E^*\right)$, so that $\sigma$ can be identified with an element of $T\left(E^*\right)$ which, moreover, satisfies $\sigma \circ \pi_n=\sigma$. Hence, $\sigma\left(\mathbf{s}_1\right) \neq \sigma\left(\mathbf{s}_2\right)$. This implies that, for each $i \in\{1, \ldots, n\}$, there exists $\mathbf{e}^* \in T((E))^*$ such that $\mathbf{e}^*\left(S\left(X_k\right)\right)=1$ if $k=i$ and 0 otherwise. This terminates the proof.