A numerated real bundle embeds in a finite trivial bundle

If $X=\mathrm{\varnothing }$, the conclusion is vacuous; assume $X$ is nonempty. For each $i$ define the $i$th component of $\mathrm{\Phi }:E\to X\times {\mathbb{R}}^{rs}$ over $x$ by $\sqrt{{\rho }_i(x)}{\tau }_i(v)$ when $x\in U_i$, and by zero when $x\mathrm{\notin }{U}_i$. The real square-root function is continuous on $[0,\mathrm{\infty })$ (away from zero, $\mathrm{\mid }\sqrt{a}-\sqrt{b}\mathrm{\mid }=\mathrm{\mid }a-b\mathrm{\mid }\mathrm{/}(\sqrt{a}+\sqrt{b})$; at zero, $0\le \sqrt{a}<\epsilon$ whenever $0\le a<{\epsilon }^2$). Because $\mathrm{supp}({\rho }_i)\subseteq U_i$, every point outside $U_i$ has a neighborhood on which ${\rho }_i=0$; hence extending the chart component by zero is continuous there, and within $U_i$ it is a product of continuous coordinate functions. Hence each component, and therefore $\mathrm{\Phi }$, is continuous. It is fiberwise linear.

If $v\in E_x$ is nonzero, the partition identity gives some $i$ with ${\rho }_i(x)>0$. Then $x\in U_i$ and [F1] makes ${\tau }_i(v)\mathrm{\ne }0$, so the $i$th component of ${\mathrm{\Phi }}_x(v)$ is nonzero. Thus each ${\mathrm{\Phi }}_x$ is injective and has rank $r$.

In local bundle coordinates, the matrix of $\mathrm{\Phi }$ has rank $r$ at every point. At any chosen $x$, some $r\times r$ minor is nonzero; continuity keeps that minor nonzero on a neighborhood. After reordering the target coordinates, the image there is the graph of a continuous matrix over ${\mathbb{R}}^r$. Thus $E^{\mathrm{\prime }}=\mathrm{\Phi }(E)$ is a rank-$r$ subbundle of the trivial bundle.

On a neighborhood with a continuous frame matrix $Q$ for $E^{\mathrm{\prime }}$, the orthogonal projection onto $E^{\mathrm{\prime }}$ is $P=Q(Q^{\mathsf{T}}Q{)}^{-1}{Q}^{\mathsf{T}}$. The Gram matrix $Q^{\mathsf{T}}Q$ is positive definite and invertible; its inverse varies continuously because each inverse entry is a cofactor divided by the nonzero determinant, so $P$ is continuous. Therefore $I-P$ has locally constant rank $rs-r$ and its image is a subbundle $F=E^{\mathrm{\prime }\perp }$. Fiberwise orthogonal decomposition gives $E^{\mathrm{\prime }}\oplus F=X\times {\mathbb{R}}^{rs}$, and $\mathrm{\Phi }$ identifies $E$ with $E^{\mathrm{\prime }}$. If $r=0$, take $E^{\mathrm{\prime }}=0$ and $F=X\times {\mathbb{R}}^0$.