(i) We can construct such a set of subspaces:

1) $\mathrm{\forall}r\in \mathbb{R}$, let: $V}_{r}=\{(x,rx)\in {\mathbb{R}}^{2}\mid x\in \mathbb{R}\$.

[Geometrically, $V}_{r$ is the line through origin of $\mathbb{R}}^{2$, os slope r.]

2) We will check that these subspaces justify assertion (i).

3) Clearly: ${V}_{r}\subseteq {\mathbb{R}}^{2}.$

4) Check that: $V}_{r$ is a proper subspace of ${\mathbb{R}}^{2}.$.

Let: $u,v\in {V}_{r},\alpha ,\beta \in \mathbb{R}$. Verify that: $\alpha u+\beta v\in {V}_{r}$

$u,v\in {V}_{r},\Rightarrow u=({x}_{1},r{x}_{1}),v=({x}_{2},r{x}_{2});$ for some $x}_{1},{x}_{2}\in \mathbb{R$

$\therefore \alpha u+\beta v=\alpha ({x}_{1},r{x}_{1})+\beta ({x}_{2},r{x}_{2})$

$=\alpha ({x}_{1},r{x}_{1})+\beta ({x}_{2},r{x}_{2})$

$=(\alpha {x}_{1},\alpha r{x}_{1})+(\beta {x}_{2},\beta r{x}_{2})$

$=(\alpha {x}_{1}+\beta {x}_{2},\alpha r{x}_{2}+\beta r{x}_{2})$

$=(\alpha {x}_{1}+\beta {x}_{2},r(\alpha {x}_{1}+\beta {x}_{2})$

$=({x}_{3},r{x}_{3})\in {V}_{r}$; with $x}_{3}=\alpha {x}_{1}+\beta {x}_{2$

So: $u,v\in {V}_{r},\alpha ,\beta \in \mathbb{R}\Rightarrow \alpha u+\beta v\in {V}_{r}$.

Thus: $V}_{r$ is a subspace of $\mathbb{R}}^{2$

To see that $V}_{r$ is non-zero, note that:

$(1,r)\in {V}_{r}$ and $(1,r)\ne (0,0)$.

To see that $V}_{r$ is proper, note that $(1,r+1)\notin {V}_{r}$:

$(1,r+1)\in {V}_{r}\Rightarrow$ (by construction of $V}_{r$) $r\cdot 1=r+1$

$\Rightarrow r=r+1$, plainly impossible.

Thus: $V}_{r$ is a non-zero, proper subspace of