It will not be necessary to use induction in this problem. One simply notes that a subspace is a subset of a vector space that is also a vector space on its own. One of the properties of being a vector space is that it is closed under addition and scalar multiplication. That is all that is being used here. I suppose if one really wanted to be rigorous one could potentially use induction I suppose.

n=1

by closure under scalar multiplication

Induction hypothesis for n=k

Assume for any .

Show . By induction hypothesis, so say .

Furthermore by closure under scalar multiplication.

Then

by closure under addition.

I think it a little extreme, but you asked for the induction. Hope that helps.