Infinite subspaces for a vector space that cannot be spanned by a single element If a vector space (over an infinite field) cannot be spanned solely by a single element, does it mean it has infinite subspaces?
I couldn't find an example that contradicts this
 A: yes
It must have two linearly independent elements $x$ and $y.$ For any $\theta \in (0,1)$ consider the set spanned by $\theta x + (1-\theta) y.$ This will be a different subspace for each $\theta.$
A: edit: It seems that Thomas is asking for is a proof that such a vector space has infinitely many subspaces. The following answer  rather regards the existence of infinite subspaces.
(First note that a vector space $V$ has infinite subspaces iff it is infinite. Maybe what you meant is infinite proper subspaces.)
You couldn't find a counter example because as others have pointed out, there is none.
Given a non zero vector $v$ of your vector space $V$ over $k$, the map $f_v: k \rightarrow V$ such that $\forall \lambda \in k, f_v(\lambda) = \lambda.v$ is injective.
This is because $k$ is a field, so for $\lambda \in k^{\times}, {\lambda}^{-1}.(\lambda.v) = ({\lambda}^{-1}\lambda).v = v \neq 0_V$, whereas ${\lambda}^{-1}.0_V + {\lambda}^{-1}v = {\lambda}^{-1}.v$ so ${\lambda}^{-1}.0_V = 0_V$.
So either $V = \{0_V\}$ or $k.v$ is an infinite proper subspace of $V$, where $v \neq 0_V$. If $V$ can't be spanned by a single element (that is $\dim_k(V) > 1$), then of course $V \neq \{0_V\}$.
Things can be different for modules over infinite rings. The axioms of modules have the property that for any two (unital) rings $A,B$ and any homomorphism $\varphi: A \rightarrow B$, if $\star_B: B \times V \rightarrow V$ is a  module external law, then the law $A \times V \rightarrow V$ given by $\lambda \star_A v := \varphi(\lambda) \star_B v$ is a module external law. An informal explanation of this fact is that those axioms don't contain the  $\neg, \Leftarrow$ and $\Rightarrow$ symbols, but you can easily check for yourself.
If the context is linear vector spaces, then $A,B$ are fields and $\varphi$ is actually a monomorphism so all we have done is restrict the external law to a subfield of $B$. 
But if $A,B$ are just rings, $\varphi$ need not be injective, and the ring structure of $A$ can be very different from that of $B$. What matters here is that $B$ can be finite without $A$ being finite. 
For instance, $(\mathbb{F}_2)^2 = \{(0,0);(1,0);(0,1);(1,1)\}$ is a finite $\mathbb{F}_2$ free module with $\dim_{\mathbb{F}_2}((\mathbb{F}_2)^2) = 2 > 1$. Using the canonical projection $\mathbb{Z} \rightarrow \mathbb{F}_2$, you get a $\mathbb{Z}$ module that can't be spanned by a single element.
For similar results with $A$ of higher cardinality, you can consider $A = \beta(E)$ the set of subsets of a non-empty set $E$. $(A,\Delta,\cap)$ is a ring (it is called a boolean ring), and given $e \in E$, the map $\varphi(S) = 1$ if $e \in S$, $\varphi(S) = 0$ otherwise is a homomorphism $A \rightarrow \mathbb{F}_2$, so you get finite $A$-modules that can't be spanned by a single element, with $|A|$ as high as you want.
