Can a linear subspace in Banach space be the union of several other subspaces? It's a well known exercise for students to prove that a linear subspace of $\mathbb{R}^n$ can't be expressed as the countable union of other subspaces. The proof is quite simple, including only the comparison of Lebesgue measure.
However, it is not true for the linear space of polynomials. It can be the union of the subspace of degree 1 polynomials, polynomials of degree $\leq 2$, degree $\le 3$, ....
By Baire's theorem, a Banach space cannot be covered by countable number of closed subspaces. So my question is whether a (Banach) vector space of uncountable dimension can be written the union of countably many proper subspaces.
PS: I know it is rude to consider union of linear subspaces--it may be more appropriate to consider sums.... Just like Goldbach, one should not try to add primes, but multiply them:)

Let me complement YCor's answer in the case where the Continuum Hypothesis fails. 
Suppose that $X$ is a Banach space with a long Schauder basis $(e_\alpha)_{\alpha < \mathfrak{c}}$ of cardinality continuum, for example, $X = \ell_p([0,1])$ for $p\in [1,\infty)$. Then every $x\in X$ is countably supported with respect to the basis. Denote by $(e^*_\alpha)_{\alpha < \mathfrak{c}}$ the coordinate functionals associated to the basis.
Bearing in mind that $\aleph_1$ is less than the continuum (we assume that CH fails), we pick $\aleph_1$ among the basis vectors, say, $e_\alpha$ for $\alpha \in A$, where $A$ has cardinality $\aleph_1$. Then,
$$X = \bigcup_{\alpha\in A} \ker e_\alpha^*,$$
because for $x\in X$ the set $\{\beta< \mathfrak{c}\colon \langle e_\beta^*, x\rangle \neq 0\}$ is countable. 
