It's the same idea as R is contractable to a point. Map f:X x R → X as f(x,y) = x, and g:X → X × R as g(x) = (x,0). Then f(g(x)) = x, and g(f(x,y)) = (x,0). To show the latter is homotopic to the identity, use h(t,(x,y)) = (x,t\*y).

You could also see it as if A is homotopy equivalent to A' and B to B' then A×B is to A'×B'. Then since R is contractable to {0} you get X × R is homotopy equivalent to X × {0}, which is homeomorphic to X.