The question is from Sheldon Axler's LADR (3rd ed.), exercise 3E, problem 14.
Problem:
Suppose $U$ is the subset of $\mathbf{F}^\infty$ defined by
$$
U = \{ (x_1, x_2, \ldots) \in \mathbf{F}^\infty: x_j \ne 0 \text{ for only finitely many } j\}.
$$
Show that $U$ is a subspace of $\mathbb{F}^\infty$ and that $\mathbb{F}^\infty/U$ is infinite dimensional.
Answering my own question:
(Showing that $U$ is a subspace is easy and skipping the proof)
Suppose $\mathbb{F}^\infty/U$ is finite dimensional and $dim \ \mathbb{F}^\infty/U = n$. We define $v_i, i = 1,...,n$ as follows: The $j$-th element of $v_i$, written as $v_i^j$ is defined as
\begin{equation} v_i^j=\begin{cases} 1, & \text{if $j = i + k(n + 1), \ k=0,1,2,...$}.\\ 0, & \text{otherwise}. \end{cases} \end{equation} So for example, $v_1 = (\underbrace{1,0,...,0,0}_{n+1 elements},\underbrace{1,0,...,0,0}_{n+1 elements},...)$
Clearly, for $\lambda_i \in \mathbb{F}, i = 1,...,n$, we have $\sum\limits_{i=1}^n \lambda_iv_i = (\lambda_1,...,\lambda_n,0,\lambda_1,...,\lambda_n,0,...)$.
$$\text{Let} \sum\limits_{i=1}^n\lambda_i(v_i + U) = (\sum\limits_{i=1}^n \lambda_iv_i) + U = 0\\ \implies \sum\limits_{i=1}^n \lambda_iv_i = (\lambda_1,...,\lambda_n,0,\lambda_1,...,\lambda_n,0,...) \in U $$
By definition of $U$ (only finitely many non-zero elements), we establish $\lambda_i = 0 \ \forall i = 1,...,n$. i.e. $v_i + U, i = 1,...,n$ are linearly independent and must be a basis (since $dim \ \mathbb{F}^\infty/U = n$).
However, now we can easily show that $v_{n+1}$ (defined the same way as other $v_i$'s) together with $(v_1, ..., v_n)$ will also be linearly independent. Hence we arrive at a contradiction and infer that $\mathbb{F}^\infty/U$ must be infinite dimensional.