Exhausting fibrations by fibrations with finite type fibers If $E \to B$ is a fibration, is it always possible to find fibrations $E_n \to B$ with maps over $B$ $E_n \to E_{n+1}$ such that $E$ is weakly equivalent to the homotopy colimit of the $E_n$'s and each $E_n$ has fibers that are weakly equivalent to finite CW complexes?
 A: Addedum:
What I wrote in the case of the universal cover of $S^1$ can be generalized to the path fibration
$$
PB \to B
$$ 
of any non-contractible connected based finite complex $B$. 
If there were a map 
$$
E_\alpha \to PB
$$
over $B$ with $E_\alpha$ non-empty, then it follows that the
map $E_\alpha \to B$ is null homotopic. This implies that the fibers
of the latter are homotopy equivalent to $E_\alpha \times \Omega B$.
But the loop space $\Omega B$ isn't the homotopy type of a finite complex if $B$ is homotopy finite (or even finitely dominated) if $B$ isn't contractible.
Second Edit:
There are clearly cases when the question has a negative answer. Here's a simple example: consider the universal cover of the circle
$$
\Bbb Z \to \Bbb R \to S^1 .
$$
Then for any map $E\to \Bbb R$ of fibrations over $S^1$ it follows that the fiber $F$ of $E\to S^1$ at the basepoint maps $\Bbb Z$-equivariantly to $\Bbb Z$. It follows that $F\to \Bbb Z$ is onto on $\pi_0$, so $F$ is never the homotopy type of a finite complex unless $E$ is empty.
Edit: 
Below is my original answer which failed to parse the question correctly (I considered the wrong notion of finiteness). 
My argument shows that $E$ is a homotopy colimit of $E_\alpha$, where $E_\alpha \to B$ is a fibration and
$E_\alpha$ is a finite complex. It doesn't show that the fibers are finite complexes.
––––––––––––original answer follows––––––––––––––––––––
The answer is yes.  Without loss in generality, we can take $B$ to be connected and based and we can assume $B=BG$ where $G$ is a topological group model for the loop space on $B$.
Then a space over $B= BG$ corresponds to a $G$-space (the Borel construction defines a functor in one direction, this functor induces an equivalence on homotopy categories of $G$-spaces and spaces over $B$, where weak equivalence in each case is defined by the forgetful functor to spaces). 
The rest of the argument is to translate your question to one about $G$-spaces and then to solve the problem in that context.
The argument goes like this:
given a $G$-space $X$, the usual procedure of killing homotopy groups enables us to write $X$ as a telescoping homotopy colimit of a diagram
$$
\{Y_\alpha\}_{\alpha\in I}
$$
of $G$-spaces which are finite free $G$-CW complexes (where a cell is of the form $D^n \times G$ (note that the indexing set $I$ can have a large cardinality.) This is almost identical to the argument which shows that an space can be written as the hocolim of finite complexes. The only difference is that we are equivariantly attaching the cells.
This suffices to give what we want, since the Borel construction of such a space is clearly weak equivalent to a finite complex.
