Now that Tom Goodwillie has basically answered the question, I feel I can undelete my non-answer. I wrote this when I had misunderstood what the OP wanted, but I feel like it's worth putting out there for people to see. The point is to show how far you can go in developing the theory without the hypotheses of Hausdorff and paracompact.

You can define vector bundles in a great level of generality. Here is the most general approach I know, which I learned from Mark Hovey. Let $B$ be a topological space. A **space over $B$** is a space $E$ and a continuous map $p:E\to B$, i.e. an element in the category $Top \downarrow B$. A **vector space over $B$** is an object $V$ of this category together with a field $F$ and maps $+:V\times V\to V, \ast: F\times V\to V, (-1):V\to V$, and $0:pt \to V$ such that a bunch of diagrams commute. These are the diagrams for associativity of $+$ and $\ast$, commutativity for $+$, distributivity, identity, inverse, etc. As an example, here is the diagram which says $x+0=x$:

$\begin{array}{ccc} V & \rightarrow & V\times pt\\\
\\\
|| & & \downarrow (id,0) \\\
\\\
|| & & V\times V\\\
\\\
|| & & \downarrow +\\\
\\\
V & = & V
\end{array}$

The product of two vector spaces $p:E\to B$ and $p':E'\to B$ is the pullback $E\times_B E'\to B$. It is an easy exercise to deduce from the axioms above that $p$ must be surjective. For all $b\in B$ define $F_b = p^{-1}(b)$. Note that for topological vector space $V$ you can form one of these vector spaces over $B$ by setting $E = V\times B$ (so all $F_b = V$) and $p$ to be projection. This is the **trivial vector space over $B$.** More generally, a vector space over $B$ is called trivial if there is an isomorphism in the category of spaces over $B$ between $E$ and $V\times B$ for some topological vector space $V$.

Given any $U\subset B$ with $i:U\hookrightarrow B$, one has a vector space over $U$, denoted $i^*(p)$, defined by $p:p^{-1}(U)\to U$. A **vector bundle over $V$** is a locally trivial vector space over $B$, i.e. for all $b$ there is a neighborhood $U$ s.t. $i^*{p}$ is trivial.

Note that none of this development required any hypotheses on $B$. As the OP mentions, the proof of homotopy invariance of the functor $X\mapsto Vect(X)$ requires paracompactness, though it's not clear to me that it requires the Hausdorff property. Even without paracompactness you can still prove $Vect(X)$ is a contravariant functor and that for any bundle $p$ over $X\times I$ there are open sets $U\times I$ covering $X\times I$ over which $p$ is trivial (by the Lebesgue covering lemma, since $I$ is a compact metric space, and ordered). However, without partitions of unity you can't compare $i_0^*p$ with $i_1^*p$ and conclude that they are the same if there's a homotopy $H$ between $f=Hi_0$ and $g=Hi_1$. It seems to me that you really can't get away with less than partitions of unity, and this is the same as paracompactness. You might be able to push the theory a little bit further without the Hausdorff hypothesis, but you'll probably need the hypothesis eventually anyway in order to say nice things.