The other answers asked you first to well-order the whole vector space (or a basis for it), and those answers are perfectly correct, but perhaps you don't like well order the whole space. So let me describe a construction that appeals directly to the Axiom of Choice.

Let V be your favorite vector space having uncountable dimension. For each countable dimension subpace W, let a_{W} be an element of V that is not in W. Such a vector exists, since W is countable dimensional and V is not, and we choose such elements by the Axiom of Choice.

Having made these choices, the rest of the construction is now completely determined. Namely, we construct a linearly ordered chain of countable dimensional spaces, whose union is uncountable dimension. Let V_{0} be the trivial subspace. If V_{α} is defined and countable dimensional, let V_{α+1} be the space spanned by V_{α} and the element a_{Vα}. If λ is a limit ordinal, let V_{λ} be the union of all earlier V_{α}. It is easy to see that { a_{Vβ} | β < α} is a basis for V_{α}. Thus, the dimension of each V_{α} is exactly the cardinality of α. In particular, if ω_{1} is the first uncountable ordinal, then V_{ω1} will have uncountable dimension, yet be the union of all V_{α} for α < ω_{1}, which all have countable dimension, as desired.

If you forbid one to use the Axiom of Choice, then it is no longer true that every vector space has a basis (since it is consistent with ZF that some vector spaces have no basis), and the concept of dimension suffers in this case. But some interesting things happen. For example, it is consistent with the failure of AC that the reals are a countable union of countable sets. R = U A_{n}, where each A_{n} is countable. (The irritating difficulty is that although each A_{n} is countable, one cannot choose the functions witnessing this uniformly, since of course R is uncountable.) But in any case, we may regard R as a vector space over Q, and if we let V_{n} be the space spanned by A_{1} U ... U A_{n}, then we can still in each case make finitely many choices to witness the countability and conclude that each V_{n} is countable dimensional, but the union of all V_{n} is all of R, which is not countable dimensional.

subsetsof $\mathbf S$. If that is so, then why not just let the chain be indexed by the countable subsets (under inclusion), with the subspace indexed by $\mathbf S'$ being its span? – L Spice Nov 17 at 9:34