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This is a late response to the post, but I noticed that the question was not answered in general.

No vector space is the finite union of proper subspaces.

EDIT: In response to my false solution, Phil Hartwig pointed out that $\mathbb{F}$$_{2}^2$ is a vector space that is the union of three proper spaces. Indeed, the "routine" induction was less routine and more nonsensical. I had fixed my proof, only to realize that my solution was much less elegant than Halmos' solution found in his Linear Algebra Problem Book. You can view the page here.

In the class of Banach spaces there is a stronger result:

If $B$ is a Banach space, then $B$ is not the countable union of proper subspaces.

This relies on the fact that a proper subspace of a topological vector space has empty interior. To appeal to your intuition in $\mathbb{R}^3$, every proper subspace (a plane or line through the origin) cannot completely contain a sphere an open ball (an open set in the usual norm topology).

Since $B$ is complete (by definition), by Baire's Theorem it is not the countable union of nowhere dense sets. Since proper subspaces are nowhere dense, $B$ is not the countable union of proper subspaces.
show/hide this revision's text 4 Deleted a false solution; provided link to a correct one.

This is a late response to the post, but I noticed that the question was not answered in general.

No vector space is the finite union of proper subspaces.

It's an easy exercise in Axler (Ch. 1, Ex. 9)

EDIT: In response to show that if $U$ and $W$ are subspaces of a vector space $V$, then $U\cup W$ is a subspace if and only if (without loss of generality) $U\subseteq W$. Since $V$ is a vector spacemy false solution, if a finite family of proper subspaces $\{U_{i}\}_{i=1}^{n}$ such Phil Hartwig pointed out that $V=\bigcup_{i=1}^{n}U_{i}$, then $\bigcup_{i=1}^{n}U_{i}$ \mathbb{F}$$_{2}^2$ is a vector space . Routine induction shows (without loss of generality) that $U_{i}\subseteq U_{n}$ for all $1\le i\le n$. But $U_{n}$ is a the union of three proper subspacespaces. Indeed, the "routine" induction was less routine and so $V\ne U_{n}$more nonsensical. I had fixed my proof, only to realize that my solution was much less elegant than Halmos' solution found in his Linear Algebra Problem Book. You can view the page here.

In the class of Banach spaces there is a stronger result:

If $B$ is a Banach space, then $B$ is not the countable union of proper subspaces.

This relies on the fact that a proper subspace of a topological vector space has empty interior. To appeal to your intuition in $\mathbb{R}^3$, every proper subspace (a plane or line through the origin) cannot completely contain a sphere (an open set in the usual norm topology).

Since $B$ is complete (by definition), by Baire's Theorem it is not the countable union of nowhere dense sets. Since proper subspaces are nowhere dense, $B$ is not the countable union of proper subspaces.
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This is a late response to the post, but I noticed that the question was not answered in general.

No vector space is the finite union of proper subspaces.

It's an easy exercise in Axler (Ch. 1, Ex. 9) to show that if $U$ and $W$ are subspaces of a vector space $V$, then $U\cup W$ is a subspace if and only if (without loss of generality) $U\subseteq W$. Since $V$ is a vector space, if a finite family of proper subspaces {$U_{i}$}${i=1}^{n}$ $\{U_{i}\}_{i=1}^{n}$ such that $V=\bigcup{i=1}^{n}U_{i}$, V=\bigcup_{i=1}^{n}U_{i}$, then $\bigcup_{i=1}^{n}U_{i}$ is a vector space. Routine induction shows (without loss of generality) that $U_{i}\subseteq U_{n}$ for all $1\le i\le n$. But $U_{n}$ is a proper subspace, and so $V\ne U_{n}$.

In the class of Banach spaces there is a stronger result:

If $B$ is a Banach space, then $B$ is not the countable union of proper subspaces.

This relies on the fact that a proper subspace of a topological vector space has empty interior. To appeal to your intuition in $\mathbb{R}^3$, every proper subspace (a plane or line through the origin) cannot completely contain a sphere (an open set in the usual norm topology).

Since $B$ is complete (by definition), by Baire's Theorem it is not the countable union of nowhere dense sets. Since proper subspaces are nowhere dense, $B$ is not the countable union of proper subspaces.
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