Complete space as a disjoint countable union of closed sets

It is a consequence of Baire's theorem that a connected, locally connected complete space cannot be written $$X = \bigcup_{n \geq 1}\ F_n$$ where the $F_n$ are nonempty, pairwise disjoint closed sets.

Does anyone know of a counter-example to this if we don't assume the space to be locally connected?

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How many sets $F_n$ do you want? Countably infinitely many? – Rasmus Jan 5 '12 at 16:01
Just a trivial remark to avoid further erroneous answers: it must be an infinite disjoint union. A finite union won't work because of connectedness. – t.b. Jan 5 '12 at 16:05
To be clear: a complete connected non-locally connected metric space that can be written as a countable union of pairwise disjoint closed sets is sought, right? – t.b. Jan 5 '12 at 16:17
@t.b. : that's right. Also one more remark: the space can't be compact, because of a theorem of Sierpinski. So if we look for some part of euclidean space it shouldn't be bounded. – timofei Jan 5 '12 at 19:47
You use the word complete, but you don't say metric space. Do you mean a complete metric space? – JDH Jan 6 '12 at 0:27

If $e_n$ are the standard unit vectors in $\ell^2$, let $F_j$ consist of line segments from $e_j$ to $(1/j) e_j + e_k$ for $1 \le k < j$ ($F_1$ is the single point $e_1$). Then the $F_j$ are disjoint, closed and connected, and their union is closed and connected.