Here's a proof of the finite dimensional case over any field (finite or infinite):
Theorem. Let $\mathbf{V}$ be a nonzero finite dimensional vector space over $\mathbf{F}$. If $\mathbf{V}$ is a union of $\kappa$ proper subspaces, then $\kappa\geq|\mathbf{F}|$.
Proof. Write $\mathbf{V}=\bigcup\limits_{k\in\kappa} W_{k}$, with $W_k$ a proper subspace of $\mathbf{V}$. By enlarging the $W_k$ as necessary, we may assume that $\dim(\mathbf{V}/W_i) = 1$ for all $i$.
The proof is by induction on $\dim(\mathbf{V})$. The result is trivially true if $\dim(\mathbf{V})=1$, because $\mathbf{V}$ is never the union of proper subspaces in this case.
For the case of $\dim(\mathbf{V})=2$, let $\{w_1\}$ be a basis for $W_1$, and let $v\notin W_1$. For each $\alpha\in \mathbf{F}$ there exists $j_{\alpha}\in\kappa$ such that $w_1+\alpha v\in W_{j_{\alpha}}$. Moreover, if $\alpha\neq\beta$, then $w_1+\alpha v$ and $w_1+\beta v$ are linearly independent, since $\{w_1,v\}$ are a basis for $\mathbf{V}$. Thus, no $W_k$ contains more than one $w_1+\alpha v$. This gives an injection from $\mathbf{F}$ to $\kappa$, proving that $\kappa\geq|\mathbf{F}|$, as required.
Assume the result holds for $n$-dimensional vector spaces, and let $\mathbf{V}$ be $(n+1)$-dimensional. Let $\{w_1,\ldots,w_n\}$ be a basis for $W_1$, and $v\notin W_1$. For each $\alpha\in\mathbf{F}$, consider the $n$-dimensional subspace $W_{\alpha}=\mathrm{span}(w_1+\alpha v,w_2,\ldots,w_n)$. If $W_{\alpha}$ is contained in some $W_k$, then $W_{\alpha}=W_k$ by dimension considerations; and if $W_{\alpha}=W_{\beta}$, then $\alpha=\beta$, for otherwise we would be able to find a nontrivial linear dependency involving $w_1,\ldots,w_n,v$. So there is again an injection from the set
$$S=\{\alpha\in\mathbf{F} \mid W_{\alpha}=W_k\text{ for some }k\in \kappa\}$$
to $\kappa$ (assuming WLOG that the $W_k$ are pairwise distinct). If the set has cardinality $|\mathbf{F}|$ we are done. Otherwise, let $\alpha_0\in\mathbf{F}\setminus S$, and look at $W_{\alpha_0}$. For each $k\in \kappa$, $W_{\alpha_0}\cap W_k\neq W_{\alpha_0}$; since
$$W_{\alpha_0} = W_{\alpha_0}\cap\mathbf{V} = W_{\alpha_0}\cap\left(\bigcup_{k\in\kappa}W_k\right) = \bigcup_{k\in\kappa}(W_{\alpha_0}\cap W_k),$$
then by the induction hypothesis we have that $\kappa\geq|\mathbf{F}|$. QED
I suspect a similar argument can be made in the infinite dimensional case; certainly, we can construct the analogous set to $S$, and if $|S|=|\mathbf{F}|$ then we are done. But if not, then $\dim(W_{\alpha})=\dim(\mathbf{V})$, so we cannot really use a reduction argument. But I think there may be a way to tweak this.
As Pete L. Clark has noted, the result does not hold in the infinite dimensional case.
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.
Best Answer
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*}