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I know that uncountably product of $\{0,2\}$ is not-metrizable, separable, compact, Hausdorff and not second countable. But what we can say about Lindelöf property? I think it is not Lindelöf, but how can I show it? Could you give me any idea?

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If by lindelof property, you mean wether it is a lindelof space, then the answer is trivially yes. Compactness implies lindelof. – Alex Becker Dec 23 '12 at 5:49
oh, yes, I have to sleep, so sorry. – ege Dec 23 '12 at 5:52
Not all uncountable products of a 2-point space are separable (you need at most continuum many copies). All such spaces are ccc though. – Henno Brandsma Dec 23 '12 at 8:20

It is obviously Lindelöf (every open cover has a countable subcover) as it is already compact (every open cover has a finite subcover). It is not hereditarily Lindelöf, nor hereditarily separable, as the subspace of all points with exactly one coordinate equal to 2 (and all others 0) is discrete and uncountable, and so not Lindelöf and not separable.

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