A vector space is finite dimensional if all of its proper subspaces are finite dimensional 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? 

Let $V$ be an infinite dimensional vector space, let $v$ be a vector of $V$. Then in particular there exists a countable subset $\{v_1,\cdots, v_n,\cdots\}$ of $V$ such that $\{v_1,\cdots, v_n,\cdots\}$ spans a subspace of $V$. We may assume $\{v_1,\cdots, v_n,\cdots\}$ is a basis, so in particular $\{v_2,\cdots, v_n,\cdots\}$ spanns a proper subspace $U$ of $V$ which is of countable dimensionality. The finite dimensional case is automatic. 
