The standard unit families $e_k \in F^{\mathbb{N}}$ form a basis of the linear subspace of eventually zero families: an explicit infinite basis, built with no choice principle

Claim 1. $E$ is nonempty: the zero family has value $0_F$ everywhere, so $N = 0$ witnesses that it lies in $E$. And $E$ is closed under the one-step expression: for $\lambda \in F$ and $u, v \in E$ with witnesses $N_1$ and $N_2$, let $N$ be the larger of the two, which exists because the order of $\mathbb{N}$ is total; then for $n \ge N$ we have $n \ge N_1$ and $n \ge N_2$, so $(\lambda u + v)(n) = \lambda u(n) + v(n) = \lambda 0_F + 0_F = 0_F$, and $N$ witnesses $\lambda u + v \in E$. So $E$ is a linear subspace of $F^{\mathbb{N}}$ by the one-step test.

Each $e_k$ lies in $E$, so $B \subseteq E$: if $n \ge \sigma(k)$ then $k < n$, hence $n \ne k$ and $e_k(n) = 0_F$, so $\sigma(k)$ is a witness.

For $x \in F^{\mathbb{N}}$ and $N \in \mathbb{N}$ put $s_N := \sum_{k<N} x(k)\,e_k$. Then $s_N(n) = x(n)$ for $n < N$ and $s_N(n) = 0_F$ for $n \ge N$. Indeed $s_N(n) = \sum_{k<N}\bigl(x(k)e_k\bigr)(n) = \sum_{k<N} x(k)\,e_k(n)$ by pointwise evaluation and pointwise scalar multiplication; the scalar list $k \mapsto x(k)e_k(n)$ has the value $x(k)0_F = 0_F$ at every $k \ne n$. If $n < N$ this list vanishes off the single index $n$, where its value is $x(n)1_F = x(n)$, so the sum is $x(n)$; if $n \ge N$ then no $k < N$ equals $n$, the list is all $0_F$, and the sum is $0_F$.

Claim 3, independence. The map $k \mapsto e_k$ is injective, since $e_k(k) = 1_F \ne 0_F = e_l(k)$ for $l \ne k$. Let $w : m \to B$ be an injective finite list and $\lambda : m \to F$ with $\sum_{j<m}\lambda_j w_j = 0$ in $F^{\mathbb{N}}$. Each $w_j$ is $e_{i_j}$ for exactly one $i_j \in \mathbb{N}$, and $j \mapsto i_j$ is injective because $w$ is. Fix $j < m$ and evaluate at $i_j$: pointwise evaluation gives $0_F = \sum_{l<m}\lambda_l\,e_{i_l}(i_j)$, and $e_{i_l}(i_j) = 0_F$ unless $i_l = i_j$, that is unless $l = j$, where it is $1_F$. So the scalar list vanishes off the single index $j$ and sums to $\lambda_j 1_F = \lambda_j$, giving $\lambda_j = 0_F$. Hence every injective finite list into $B$ is independent, that is $B$ is linearly independent.

$B \approx \mathbb{N}$: the map $k \mapsto e_k$ is injective by step 1.4 and its image is $B$ by definition, so it is a bijection $\mathbb{N} \to B$.

Claim 2. Each $e_k$ lies in $E$ and $E$ is a linear subspace, so $\operatorname{span}(B) \subseteq E$ by minimality of the span. Conversely let $x \in E$ with witness $N$; then $x$ and $s_N$ of step 1.3 agree at every $n$, since for $n < N$ both take the value $x(n)$ and for $n \ge N$ both take the value $0_F$, so $x = s_N = \sum_{k<N} x(k)e_k$, a linear combination of elements of $B$. Hence $E \subseteq \operatorname{span}(B)$.

Claim 4. Suppose $E$ had a finite basis $C$, say with $p$ elements. Then $C$ is a spanning set of $E$ with $p$ elements, so no linearly independent subset of $E$ is equinumerous with $\mathbb{N}$. But $B \subseteq E$ is linearly independent by step 1.4 and $B \approx \mathbb{N}$ by step 2.1. So no finite basis exists and $E$ is infinite-dimensional over $F$.

Claim 3, that $B$ is a basis of $E$. By step 1.4 the set $B$ is linearly independent and by step 2.2 it spans $E$; independence and spans computed in the linear subspace $E$ agree with those computed in $F^{\mathbb{N}}$, so $B$ is a basis of the vector space $E$.