Proving that $\mathbb{F}^\infty$ is infinite-dimensional.

Here's a really simple proof. Let $f\colon \mathbb{F}^{\infty}\rightarrow \mathbb{F}^{\infty}$ be given by $f(x_0,x_1,x_2,\ldots)=(x_1,x_2,\ldots)$. It's easy to see that $f$ is linear. Note that $\mbox{Im}\:f =\mathbb{F}^{\infty}$ but $\ker f=\{(x_0,0,0,\ldots)\mid x_0\in\mathbb{F}\}\cong\mathbb{F}$ and so we have a surjective linear map from a vector space to itself which is not injective. This is not possible for finite dimensional vector spaces by the rank nullity theorem and so $\mathbb{F}^{\infty}$ is infinite dimensional.

An indirect proof can be simpler than the one you sketch.

Suppose that $K^\infty$ (or $K^{\mathbb N}$ -- that doesn't matter here) has finite dimension $n$. We know already (I hope) that in a finite-dimensional vector space no linearly independent set can have more members than the dimension. But $\{\mathbf e_1,\mathbf e_2,\ldots,\mathbf e_n,\mathbf e_{n+1}\}$ is clearly a linearly independent set of size $n+1$, so $n$ can't have been the dimension after all.