(a) Prove that the only subspaces of $\Bbb{R}$ are $\Bbb{R}$ and the zero subspace.
(b) Prove that a subspace of $\Bbb{R}^2$ is $\Bbb{R}^2$, or the zero subspace, or consist of all scalar multiples of some fixed vector in $\Bbb{R}^2$.
My attempt: (a) Assume towards contradiction, $\exists A$ such that $A\neq \{0\},\Bbb{R}$ and $A\leq \Bbb{R}$. Since $A\neq \{0\}$, $\exists a\in A$ such that $a\neq 0$. By axiom of multiplication in fields, $\exists \frac{1}{a}\in \Bbb{R}$ such that $a\cdot \frac{1}{a}=1$. Since $A\neq \Bbb{R}$, $\exists r\in \Bbb{R}$ such that $r\notin A$. Define $p=\frac{1}{a}\cdot r\in \Bbb{R}$. Since $A\leq \Bbb{R}$, $p\cdot a=(\frac{1}{a}\cdot r)\cdot a=r\in A$. Thus we reach contradiction. Is this proof correct?
(b) Claim: $L_x=\{c\cdot x|c\in \Bbb{R}\}\leq \Bbb{R}^2$, $\forall x\in \Bbb{R}^2$. Proof: let $x\in \Bbb{R}^2$. (1) $0_v=0\cdot x\in L_x$. (2) $c_1\cdot x+c_2\cdot x=(c_1+c_2)\cdot x\in L_x$, since $c_1+c_2\in \Bbb{R}$. (3) $c’\cdot(c\cdot x)=(c’\cdot c)\cdot x\in L_x$, since $c’\cdot c\in \Bbb{R}$. Hence $L_x\leq \Bbb{R}^2$, $\forall x\in \Bbb{R}^2$. How to prove these are the only subspaces of $\Bbb{R}^2$? I tryed using contradiction but made no progress. I haven’t yet studied matrix theory.