Forms on vector bundles: vertically compactly supported

Definition: Let $$\pi:V \rightarrow M$$ be a vector bundle. $$\Omega^p_{cv}(V)$$ is the sections of $$p$$ form on $$V$$, such that $$\pi^{-1}(K) \cap supp \, (w)$$ for all $$K \subseteq M$$ compact.

Definition: $$\Omega^p_c(V)$$ is the set of sections $$p$$ forms on $$V$$ such that its support is compact.

It is claimed that

$$\Omega^p_c(V) = \Omega^p_{cv}(V)$$ when $$M$$ is compact.

How is this true? $$\subseteq$$ is clear.

Let $$\omega$$ be a form on $$V$$ which is compactly supported on each fiber $$V_x$$. Consider $$\text{supp}(\omega) \subset V$$; this is a closed subset of $$V$$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.
Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $$X \subset V$$ is closed, with the property that each $$X \cap V_x$$ is compact, then we may define a function $$\sigma: M \to \Bbb R_{\geq 0}$$ given by sending $$\sigma(x)$$ to the largest $$\|v\|$$ of any $$v \in X \cap V_x$$. This is well-defined because $$X \cap V_x$$ is compact. The map $$\sigma$$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $$X \cap V_x$$ are 'close', and hence should have 'nearby' size bounds.)
Thus $$\sigma$$ is a continuous map from a compact space to $$\Bbb R$$, and hence has an absolute maximum, say $$L$$. Thus the set $$X$$ is contained in the closed $$L$$-disc bundle of $$V$$, which is a compact space (given any sequence $$(m_n, v_n)$$ of norm bounded by $$L$$, choose a subsequence so that $$m_n$$ is convergent to $$m$$; working on a chart to identify all nearby fibers with $$V_x$$, use compactness of the $$L$$-disc ball to see that a further subsequence may be chosen so that $$v_n$$ converges). $$X$$, being a closed subspace of a compact Hausdorff space, is compact.
So any form with compact vertical support on a vector bundle $$V$$ over compact base $$M$$ has compact support, as desired.