What's wrong with compact-open topology on the space of maps? Given a smooth vector bundle $E$ with non-compact base, let
$\Gamma(E)$ be the space of $C^\infty$ sections equipped with compact-open $C^\infty$-topology. 


*

*I have heard that $\Gamma(E)$ is not locally-contractible. Why not?

*Is $\Gamma(E)$ contractible? Visibly any section can be joined to the zero section by "straight line", doesn't this prove that $\Gamma(E)$ is contractible?

*Is it true that every convex subset of $\Gamma(E)$ is contractible? The argument of 2 seems to apply, but then it seems plausible that each section has an arbitrary small convex neighborhood, contradicting 1.
CLARIFICATION: One source of "rumor 1" is the book "The Convenient Setting of Global Analysis" freely available 
 here. On page
429 one reads: "Unfortunately, for non-compact $M$, the space $C^\infty(M, N)$ is not locally contractible in the compact-open $C^\infty$-topology". Another source is the discussion in Hirsch's book in the beginning of Chapter 2, which says "It can be shown that
$C^\infty(M, N)$ has very nice features, e.g. it has a complete metric, and a countable base; if $M$ is compact, it is locally contractible and $C^r(M, \mathbb R^n)$ is a Banach space for $2\le r<\infty$".
Thus I assumeed that in general, if $M$ is non-compact, then the space $\Gamma(E)$ is not (or maybe just need not be?) locally contractible. Also I am uncertain whether $C^\infty(M, \mathbb R^n)$ or $\Gamma(E)$ is a topological vector space, is it really? There seems to be a sequence of semi-norms giving these spaces a structure of Frechet spaces, but then they must be locally convex, hence locally contractible. Obviously, I am missing something. 
In response to comments I ask a more specific question. 
Question. Let $T_{r,s}(M)$ denote the space of $C^\infty$-smooth $(r,s)$-tensors on a connected non-compact $C^\infty$ manifold $M$. For $k$ with $2\le k\le \infty$, give $T_{r,s}(M)$ the weak $C^k$-topology as in Hirsch's book (roughly for $k$ finite we require that given $\epsilon>0$ and compact subset $K$ all derivatives up to $k$ are $\epsilon$-close over $K$, and for $k=\infty$ we take the union of all $C^k$-topologies for all finite $k$ under the inclusions $C^\infty\to C^k$). Now I ask
Is  $T_{r,s}(M)$ a Fréchet space with respect to the weak $C^k$-topology?
In particular, I want to conclude that $T_{r,s}(M)$ is locally contractible, and any convex subset of $T_{r,s}(M)$ is contractible; I think Fréchet spaces must have this property.
 A: Are you sure it is about vector bundles?
Any topological vector space is contractible. Any locally convex topological vector space is locally contractible. This is straightforward from the definitions. The compact-open $C^\infty$ topology is defined by a family of semi-norms and hence locally convex.
In the case of arbitrary fiber bundles, the space of sections can indeed fail to be locally contractible. For example, consider the trivial bundle $\mathbb N\times\{0,1\}$ over $\mathbb N$ (they are 0-dimensional manifolds). The space of sections is not locally contractible (in fact, homeomorphic to the Cantor set). You can build connected examples as well, e.g. consider the trivial circle bundle over a surface with infinitely many handles.
A: (For the more specific question)
Yes for $k = \infty$ if $M$ can be exhausted by a countable number of compact sets, no otherwise.  However, the failure is due to a lack of completeness (for $k \ne \infty$) or size issues (if $M$ can't be exhausted) rather than anything else and thus the local contractibility still holds.  Indeed, contractibility holds simply by contracting the vector bundle itself down to the image of the zero section.
The reason is due to the fact that the topology can be described by a family of semi-norms, as Sergei indicates in his comment to his earlier answer, so you get a locally convex topological vector space.  I recommend that you read about these spaces; Schaefer's book is a good place to start (as in Jarchow's but that doesn't seem to be available any more).
(In particular, be wary of saying "union of all $C^k$-topologies"; actually you are taking a projective limit here which means that the space has a $0$-neighbourhood base which is a union of the $0$-neigbourhood bases from each of the $C^k$ topologies, but that doesn't mean that the final topology is the union of all of the $C^k$ topologies.  Simply take a set of point $x_n$ that are "far apart" and put a set $U_n$ about each one so that $U_n$ is open in the $C^n$-topology but not in $C^{n-1}$.  Then $\bigcup U_n$ is open in the $C^\infty$ topology but not in any of the $C^n$-topolgies.)
