Problem
Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbb{R}^2$.
Solution
Consider $U^2 = \mathbb{Z}^2$, this is clearly a subset of $\mathbb{R}^2$ and is also closed under addition: $a, b \in \mathbb{Z}^2$ implies $a + b \in \mathbb{Z}^2$, also $a \in \mathbb{Z}^2 \to -a \in \mathbb{Z}^2$ since $\mathbb{Z}$ is closed under addition and negation.
But $\mathbb{Z}^2$ is not a subspace over $\mathbb{R}$ since $(1,1) \in \mathbb{Z}^2$, but $\frac{1}{2} (1,1)$ $ \cancel{\in} $ $ \mathbb{Z}^2$. That is, $\mathbb{Z}^2$ is not closed under multiplication by real scalars and is therefore not a subspace of $\mathbb{R}^2$.