CW complexes and paracompactness It seems like when we assume "niceness" in homotopy theory we assume that $X$ has the homotopy type of a CW complex, and in fiber bundle theory we assume that $X$ is paracompact. How do these two interact? Is any space with the homotopy type of a CW complex paracompact? (In particular, is $I^I$ paracompact?)
(CW complexes are always paracompact and Hausdorff. According to Milnor (http://www.jstor.org/stable/1993204) a paracompact space that is "equi locally convex" will have the homotopy type of a CW complex. Also according to that paper, if $X$ has the homotopy type of a CW complex and $K$ is actually a finite complex then $X^K$ has the homotopy type of a CW complex.)
 A: For the fact that CW complexes are paracompact, I think that the proof in this book is better than I have seen elsewhere:


\bib{frpi:cst}{book}{
    author={Fritsch, Rudolf},
    author={Piccinini, Renzo~A.},
     title={Cellular structures in topology},
    series={Cambridge studies in advanced mathematics},
 publisher={Cambridge University Press},
      date={1990},
    volume={19},
}

A: $I^I$ is paracompact. It is a theorem of O'Meara that for $X$ a separable metric space, and $Y$ a metric space, then $Y^X$ - with the compact-open topology - is paracompact. $I$ is certainly a separable metric space, so the result holds.
As to your first question, I doubt that every space of the homotopy type of a CW-complex is paracompact (but this is intuition only). Something like $\mathbb{R}^{\aleph_2}$ or a similarly large-dimensional, non-metrizable topological vector space might do the trick, as it is contractible, hence the homotopy type of a CW-complex.
Edit: For any uncountable index set $J$, consider $\mathbb{R}^J$ in the product topology. It isn't normal, so isn't paracompact. 
