Answer
See below
Work Step by Step
Assume that $x$ be the vector in $S$
We can write set $S$ as $S=\{u\in R^2:x=(a,-\frac{2}{3}a),a \in R\}$
A1. Let $x=(a,-\frac{2}{3}a)\\
y=(b,-\frac{2}{3}b)\in S$
Obtain $x+y=(a,-\frac{2}{3}a)+(b,-\frac{2}{3}b)\\=(a+b,-\frac{2}{3}a-\frac{2}{3}b)\\=(a+b,-\frac{2}{3}(a+b))$
Coordinates of sum $x+y$ are of the form of vectors from $S$, hence set $S$ is closed under addition.
A2. Let $(a,-\frac{2}{3}a) \in S$
and $k \in R$
Obtain: $kx=k(a,-\frac{2}{3}a)=(ka,-\frac{2}{3}ka)$
Coordinates of product $kx$ are of the form of vectors from $S$, hence set $S$ is closed under scalar multiplication.
Since $S$ is nonempty subset of $R^2$ and $S$ is closed under addition and scalar multiplication, $S$ is subspace of $R^2$
