To expand and correct my comment...
Let $Top$ be the category of all topological spaces, $PHTop$ the category paracompact Hausdorff spaces. Let $open$ denote the Grothendieck topology of open covers on either of these categories and $num$ the Grothendieck topology of open covers for which there exists a subordinate partition of unity. Then there are morphisms of sites $Top_{num} \to Top_{open}$ and $PHTop_{num} \to PHTop_{open}$. By definition, the latter is an equivalence. But the former is not: there are genuinely more general open covers.
One can repeat this story for the small sites associated to a fixed space.
Now the stack $Bun_G$ of principal $G$-bundles on any of these sites is presented by the one-object topological groupoid $\mathbf{B}G = G\rightrightarrows *$. Any principal bundle $P\to X$ gives a map of stacks $X\to Bun_G$. If $P$ trivialises over $U \to X$, $U = \coprod_i U_i$ an open cover, then there is a span of topological groupoids $X \leftarrow \check{C}(U) \to \mathbf{B}G$. Here $\check{C}(U)$ is the topological groupoid with object space $U$ and arrow space $U\times_X U$. None of this is particularly specific to the case at hand. Since we are dealing with topological groupoids, though, we can geometrically realise and get a span of spaces $X \leftarrow B\check{C}(U) \to BG$, where now $BG$ (plain $B$!) is Segal's construction of the classifying space of $G$.
However, for $U$ an open cover with a subordinary partition of unity, the map $X \leftarrow B\check{C}(U)$ is a homotopy equivalence (a result of Segal), whereas in general it is a weak homotopy equivalence. So for numerable bundles, where local triviality is with respect to covers with partitions of unity, you get a map $X\to BG$ and this is well-defined up to homotopy. For arbitrary bundles on non-paracompact Hausdorff spaces, there is only a morphism in the homotopy category, a span with backwards-pointing leg a weak homotopy equivalence.
Additional to this, the universal $G$-bundle on $BG$ is already numerable, so any bundle gotten by pulling it back is numerable, so this construction also goes in the opposite direction: one cannot get a non-numerable bundle via pulling back along any $X\to BG$. The same goes for Milnor's construction of $BG$ as an infinite join.
So if you want the topological/homotopy version of classifying theory of principal bundles (using homotopy classes of maps) to agree with the stack-theoretic version, you'd better only use numerable bundles. Which is to say, you need to use the site $Top_{num}$. When restricted to paracompact Hausdorff spaces this the same as arbitrary open covers, but in general this is a real restriction that affects the theorem about classification of bundles.
As an additional comment, I think an example of a non-numerable bundle is the frame bundle $F(L)$ (or, equivalently, the tangent bundle $TL$) of the long line $L$. You can think of $L$ as a manifold in the sense of being locally Euclidean and Hausdorff, but isn't metrizable, hence not paracompact. It is connected, path connected and simply-connected. As a result, $F(L)$ is not trivial, despite all this (otherwise $L$ would be metrizable).
Added Ok, so $TL$ (and so $F(L)$) can't be numerable, otherwise one could define a Riemannian metric on $TL$, and hence a metric on $L$ giving its topology.
And so this gives another good point why you want numerable bundles in general: if you want connections to exist, or metrics, or indeed any type of geometric information that can be patched together using a partition of unity, then bundles need to trivialise over a numerable cover. A numerable bundle over a non-paracompact Hausdorff space does always admit connections; a numerable vector bundle admits metrics and so on.