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.)
Best Answer
$I^I$ 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.