Your example is a vector space of dimension 2, so the only proper subspaces are those of dimensions 0 and 1. You have already accounted for dimension 0. A vector space of dimension 1 consists of a single nonzero vector and all of its scalar multiples. So: pick a nonzero vector, gather all of its scalar multiples, there's a proper subspace. Then pick a vector not in that subspace, and repeat the exercise. Repeat until you have there are no more nonzero vectors left out, and you have all the proper subspaces.
The answer is yes; if $V$ is a finite vector space (so over a finite field and of finite dimension), then it has only finitely many elements. Let $\langle v\rangle$ denote the span of a vector $v\in V$. Then clearly
$$V=\bigcup_{v\in V}\langle v\rangle,$$
because $v\in\langle v\rangle$ for every $v\in V$, and the union is finite because $V$ is finite. Hence $V$ is the union of finitely many proper subspaces (if $\dim V>1$, otherwise the subspaces aren't proper).
For a very concrete example, consider $\Bbb{F}_2^2$, a $2$-dimensional vector space over the finite field $\Bbb{F}_2$ of two elements. Then
\begin{eqnarray*}
\Bbb{F}_2^2&=&\{(0,0),(1,0),(0,1),(1,1)\}\\
\bigcup_{v\in\Bbb{F}_2^2}\langle v\rangle&=&\{(0,0)\}\cup\{(0,0),(1,0)\}\cup\{(0,0),(0,1)\}\cup\{(0,0),(1,1)\}.
\end{eqnarray*}
Best Answer
Hint: If $(x_1,x_2,\ldots)\in V$ then either $x_1=0$ or there exists $c\in K$ with $x_2=cx_1$.