Why does the following set have no subspaces but {0} and itself? Here's the statement:

The following set, $V$, only has subspaces $\{0\}$ and $V$.
      $$V=\{f(t) \colon \mathbb R \to \mathbb  R \mid f'(t) = k\cdot f(t) \text{ where } k \text{ is a constant}\}$$ 

I'm having trouble understanding why there are no other subspaces.  Why is this the case here?  Examples are welcome.  I provided a definition of "subspace" in the comments.

HINT $\rm\displaystyle\ \begin{align} f{\:'} &=\ \rm k\ f \\ \rm \:\ g' &=\ \rm k\ g \end{align}\ \Rightarrow\ \dfrac{f{\:'}}f\: =\: \dfrac{g'}g\: \iff\: \bigg(\!\!\dfrac{g}f\bigg)' =\ 0\ \iff \ g\: =\: c\ f,\ \ \ c'\: =\ 0,\ $ i.e. $\rm\ c\:$  "constant".
This is a special case of the the Wronskian test for linear dependence.
