A quite verbal proof that if V is finite dimensional then S is also....

[WWGD]

Okay, but one problem is that the vectors in the basis for $V$ may not be in the subspace $S$ at all. Maybe starting with a basis for $V$ is not the right approach?

You argued in #4, that starting with a basis for $V$ is not the right approach. Not that starting with a basis for $S$, which is what I suggested, is not the right approach. At least I did not see it elsewhere. But yes, I guess then your approach is taking a nontrivial combination of n+1 basis vectors in S, which are necessarily dependent , as they live in V, so we backtrack to S and get a nontrivial combination that equals 0.

 

[PeroK]

You argued in #4, that starting with a basis for $V$ is not the right approach. Not that starting with a basis for $S$, which is what I suggested, is not the right approach.

And I argued in post #16 that starting with a basis for $S$ was not the right approach either:

I made no assumption that S was finite dimensional. I simply looked for $n + 1$ independent vectors in $S$. I avoided the potential tangle created by looking for a basis for $S$ or a set that spans $S$.

 

[PeroK]

... the tangle is that you have to deal with the possibility that the basis or spanning set is infinite. And, given that we need to prove that $S$ is finite dimensional, this makes things unnecessarily clumsy.

Whereas, simply taking a finite set of $n + 1$ independent vectors in $S$ is simple and elegant, as has already been established!

 

[WWGD]

And I argued in post #16 that starting with a basis for $S$ was not the right approach either:

I made no assumption that S was finite dimensional. I simply looked for $n + 1$ independent vectors in $S$. I avoided the potential tangle created by looking for a basis for $S$ or a set that spans $S$.

And I agree with your argument on cardinality. I offered a similar one awhile back in Stack Echange , was downvoted and told I must use induction in the length of segments $[n]:=\{1,2,..n\}$ by hard core set theorists. I bought into it since they know way more about set theory than I do. Not sure what their quibble was. Not disagreeing with your argument, just recounting what some expect as a proof.

 

[PeroK]

And I agree with your argument on cardinality. I offered a similar one awhile back in Stack Echange , was downvoted and told I must use induction in the length of segments $[n]:=\{1,2,..n\}$ by hard core set theorists. I bought into it since they know way more about set theory than I do. Not sure what their quibble was. Not disagreeing with your argument, just recounting what some expect as a proof.

This argument arises on here as well from time to time. My position is simple: in general a university maths degree does not begin with a year of hard-core set theory. We have to assume, therefore, that a subject like linear algebra can be taught without descending into set theory at every turn.

In any case, the OP is trying to learn Linear Algebra, not the intricacies of the foundations of mathematics.