TITLE: Does every vector bundle allow a finite trivialization cover?
QUESTION [21 upvotes]: Suppose there is a vector bundle  (smooth, with constant rank finite-dimensional fibres) over a (smooth, second-countable, Hausdorff, not necessarily connected) manifold $B$ of dimension $n$.
(a) Is it true, that the manifold B can be covered by a finite number of sets $U_1,\dots,U_N$ s.t. the vector bundle, restricted to $U_i$, is isomorphic to a trivial one for every $i=1,\dots,N$?
(b) If yes, can $N$ be taken to be $n+1$?
P.S. Some observations:

It's proven in the book by Milnor and Stasheff, that every bundle allows a countable locally finite trivialization cover.
Part (a) is obviously trivial for compact manifolds.
It seems, that (b) is true if $B$ is an $n$-dimensional CW-complex. Proof: Denote with $B_k$ union of cells of dimension $0,\dots,k$. Prove by induction in $k$, that there are subsets $U_0,\dots,U_k$ of $B$, which cover $B_k$, s.t. the restriction of the bundle to each of them is trivializable. Start with case $k=0$: construct contractible neighbourhoods of each 0-cell, which do not intersect with each other. Take there union. Now to prove the claim for the next value of $k$ it is enough to construct a contractible non-intersecting neighbourhoods of each $X_\alpha=e_\alpha\setminus (U_0\cup \dots\cup U_{k-1})$. Call the desired neighbourhood with $V_\alpha$. First, note, that $X_\alpha$ is closed in $e_\alpha^k$ and doesn't intersect with its boundary $\partial e_\alpha^k$, so we can find its neighbourhood in $e_\alpha^k$, which doesn't intersect with $\partial e_\alpha^k$. This set is our candidate for $V_\alpha\cap B_k$. Extending it to an open set in $B_{k+1}$ can be done cell by cell: interpreting $e^{k+1}_\beta$ as a unit ball with the center in the origin, we can write every its point as $r\theta$, where $\theta\in S^k$ and $r\in [0,1]$. We include $r\theta$ in $V_\alpha\cap B_{k+1}$ iff $\theta$ is already there and $r>0.99$. Repeating this procedure we extend it to $B$.

Edit: Open sets $U_1,\dots, U_N$ are assumed to be open. I don't ask them to be connected.

REPLY [4 votes]: (a) and (b) are also true for topological manifolds of dimension $n$, see pages 17-21 of 

MR0336650  Greub, Werner; Halperin, Stephen; Vanstone, Ray: Connections, curvature, and cohomology. Vol. I: De Rham cohomology of manifolds and vector bundles. Pure and Applied Mathematics, Vol. 47. Academic Press, New York-London, 1972.

REPLY [3 votes]: I wonder that this was not said before.
Take a triangulation of your manifold. Choose disjoint open balls around each 0-cell in the triangulation and set the union to be $U_0$ (a ball means here something diffeomorphic to an open ball). 
For each 1-cell, choose an open ball such that together with the balls chosen before around the corners, the 1-cell is completely covered, and choose those balls to be disjoint. Take the union of those to be $U_1$.
Keep on going in the same way, until you reach the $n$-cells. 
Now you have $n+1$ sets $U_0, \dots, U_n$, each consisting of disjoint unions of open balls and such that $U_0 \cup \dots \cup U_j$ covers all cells of dimension less or equal to $j$. In particular, the union of all these sets covers your manifold.
Your vector bundle must be trivial on each of the sets $U_j$, as each are disjoint unions of open balls.