Problem
Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbb{R}^2$.
Solution
With inspiration from problem 6 we construct the following subset of $\mathbb{R}^2$: $\lbrace (x,y) \in \mathbb{R}^2 | x^2 = y^2 \rbrace$. Then if $(x,y)$ is in the subset $\lambda (x,y) = (\lambda x, \lambda y)$ has the property that $(\lambda x)^2 = \lambda^2 (x^2) = \lambda^2 (y^2) = (\lambda y)^2$, that is - the subset is closed under scalar multiplication.
However $(-1,1)$ and $(1,1)$ are both in the subset because $(-1)^2 = 1^2$, but $(-1,1) + (1,1) = (0,2)$ is not in the subset because $0^2 \neq 2^2$, therefore the constructed subset of $\mathbb{R}^2$ is not closed under addition and is therefore not a subspace of $\mathbb{R}^2$.