Prove $V$ is infinite dimensional if and only if there is a sequence $v_1,v_2,\dots$ of vectors in $V$ such that $v_1,\dots,v_m$ is linearly independent for every positive integer $m$.

Recall definition 2.15:

A vector space is called infinite-dimensional if it is not finite-dimensional.

Therefor we must show $V$ is not finite dimensional, ie. that no finite list of vectors spans $V$.

From 2.23 (length of spanning list) $\ge$ (length of linearly independent list), since we can find linearly independent lists of any length no spanning list can exist. More precisely for any spanning list of length $n$, the list $v_1,\dots,v_{n+1}$ is linearly independent, contradicting the given list being spanning. Thus $V$ is infinite dimensional

Now suppose $V$ is infinite dimensional, let $v_1 \in V$, $v_2 \notin \text{span}(v_1)$ and in general $v_k \notin \text{span}(v_1,\dots,v_{k-1})$ (we can always find $v_k$ since no finite list spans $V$) then we have our sequence $v_1,v_2,\dots$ as desired.