The proof above is from Axler's Linear Algebra Done Right.
However, I'm not sure why Axler doesn't need to state that a linearly independent list of vectors can span all vector spaces. Or that a vector in the span of other vectors can be taken out and still have the same span.
Otherwise, can't I argue that the process doesn't necessarily terminate because there are other vectors in $U$ that have not been included in the span?
Best Answer
You are constructing a list consisting of vectors from $U$ that is linearly independent. Since every linearly independent list in $U$ is linearly independent in $V$, this list that you are constructing cannot be any longer than the basis of $V$(by theorem $2.23$ of the book). Thus, the process terminates. This termination of the process implies that $U$ is finite dimensional as if $U$ were infinite-dimensional, then this process would never stop as no finite list spans an infinite dimensional vector space. Have a look at theorem $2.23$ of the book, it is a very important theorem.