My question is the title, but let me be more specific: for schemes $X$ and $Y$ over $S$, with at least one non-separated over $S$, is it true that the fibered product $X\times\_S Y$ is also not separated over $S$?

My instinct says "no" because of the remarks at the top of page 95 in Hartshorne: "The rough idea is that in order for a schemes $X$ to be separated, it should not contain any subscheme which looks like a curve with a doubled point..." So if $X$ is not separated, it contains such a subscheme, and thus so should the product $X\times\_S Y$. 

I tried proving the claim using the valuative criterion, and I think I was able to do it. Unfortunately, I don't know how to typeset diagrams here (it seems xymatrix is not supported), but the idea is to take 2 of the different morphisms from $T$ to $X$ (using the notation of Hartshorne here in which $T = \text{Spec}(R)$ for a valuation ring $R$) and make them morphisms to the product. Unless I overlooked something, the proof was pretty simple. 

Is all this correct or are there counter-examples out there? 

**EDIT**: As pointed out by Martin Brandenburg below, in general, the product of non-separated schemes could be separated. However, I think it is true that if the schemes are all over some field $k$ then non-separated-ness is preserved under taking products. Is this true, and if so, how would one prove it?