TITLE: A vector space is finite dimensional if all of its proper subspaces are finite dimensional
QUESTION [11 upvotes]: Suppose every proper subspace $(\neq V)$ of a vector space $V$ is finite dimensional. Prove that $V$ is finite dimensional. 
This question just popped into my head when I was reading about inner product space, so I can't guarantee how legitimate the question is. 
My try: 
Assume V is an inner product space. Take any proper subspace $U$ of $V$. Then $V=U\oplus U^{\bot}$. Both $U$ and $U^{\bot}$ are finite dimensional. So they both have a basis of finite dimension and we will be done. 
So for the inner product space $V$ the statement holds true. 
What about other cases? Does the statement hold true for every vector space? Does there exist a characteristic to identify vector spaces with this property?

REPLY [5 votes]: Take a subspace $U$ of codimension $1$ (why exits it?), i.e. $\dim V/U = 1$. Applying the First Isomorphism Theorem to the the projection
$$\pi:V\longrightarrow V/U,$$
we have
$$\dim V = \dim(\ker\pi) + \dim (V/U).$$
But by hypothesis $\dim(\ker\pi)<\infty$.