In Sheldon Axler's "Linear Algebra Done Right" 3rd edtion Page 36 he worte:Proof of every subspaces of a finite-dimensional vector space is finite-dimensional
The question is: I do not understand the last sentence"Thus the process eventually terminates, which means that U is finite-dimensional". So far I can understand that the subspace $U$ has limited length of independent vector list, but how to reach from here to the conclusion that subspace $U$ is finite dimensional? By 2.23 we see that the length of spanning list of vectors can be more than(rather than less than) the finite list vetors' length. Thus finite length of independent vectors list does not imply finite length of spanning list!
My proof: By the definition of subspace, we see that $U$ is a subset of $V$(with some other restrictions). Then, for every $u\in U$, $u\in V$, implying $u\in span(v_1,v_2,...,v_m)$ for $V= span (v_1,v_2,...,v_m)$ (Here I used the difinition 2.10). QED.
2.23 Length of linearly independent list
length of spanning list In a ﬁnite-dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors.
2.10 Deﬁnition ﬁnite-dimensional vectors pace
A vector space is called ﬁnite-dimensional if some list of vectors in it spans the space.
Please verify why Professor Axler's proof is valid and point out any mistakes I've made in my understanding or proof if there are some, thanks in advance!