Problem
Show that the subspaces of $\mathbb{R}^2$ are precisely $\lbrace 0 \rbrace, \mathbb{R}^2$, and all lines through the origin.
Solution
As per theorem 2.38, we must have that any subspace $U$ of $R^2$ has a basis of length $0$, $1$ or $2$. The case when a basis of $U$ has length $0$ is exactly when $U = \lbrace 0 \rbrace$, when $\dim U = 2 = \dim \mathbb{R}^2$ we must have $U = \mathbb{R}^2$ by P2C1.
When $\dim U = 1$, a basis of $U$ consist of a single element $v = (x,y)$ in $\mathbb{R}^2$, for $x,y \in \mathbb{R}$, implying that $U = \textrm{span}((x,y))$ i.e. $U$ is the set of all vectors that can be written as $\lambda (x,y)$ for $\lambda \in \mathbb{R}$. This is exactly all lines through the origin.