Let $X$ be a complex manifold equipped with a smooth hermitian metric $h$. We can define a sub-fibration $B \to X$ of the tangent bundle $T_X$ by requiring that the fiber over a point be the unit ball in $T_X$ at that point, i.e.
$ B_x = \{ \xi \in T_{X,x} \, | \, \| \xi \|_{h(x)} = 1 \}. $
I want to see that if $X$ is compact, then the fibration $B$ is compact in the total space of $T_X$. This is just a global version of the usual fact that the unit ball in a normed finite dimensional vector space is compact.
I can prove this by using the (horrible) definition of $T_X$ as the disjoint union of the stalks $T_{X,x}$ modulo an equivalence relation. As each $B_x$ is compact, their disjoint union is compact in the disjoint union of the stalks by Tychonoff, and dividing by the equivalence relation is a continuous map so $B$ is compact in the total space of $T_X$.
Isn't there a nicer way to see this? I ask both because I really don't like the definition of the tangent space as a disjoint union of stalks, and because I have to prove a similar lemma for the relative tangent space associated to deformations of a manifold $X$. A similar proof works in that case, but it is quite dirty.