Example of a subset of $\mathbb{R}^2$ that is closed under vector addition, but not closed under scalar multiplication?

$S=\{(r,s)\in\mathbb{R}^2:r,s\in\mathbb{Q}\}$ works as well.

The set $\{(x,y): x\ge0, y\ge0\}$ is closed under addition, but not under scalar multiplication, since $-1\cdot(1,1)=(-1,-1)$, for example.

The subset $S=\{(n,0)\in\mathbb{R}^2:n\in\mathbb{Z}\}$ is closed under addition but not scalar multiplication, e.g., $\frac{1}{2}\cdot (1,0)=(\frac{1}{2},0)\notin S$.

Let $\{v_1, v_2, \ldots\}$ be any countable subset of $\Bbb R^2$ whose elements are not all 0. Then $$\left\{\sum_{i=1}^\infty z_iv_i \middle\lvert z_i\in \Bbb Z\right\}$$ is closed under addition. But it is a countable set, and any set (except {0}) that is closed under scalar multiplication is uncountable.