= Solution
Work with real <vector bundles>; for complex bundles replace the <inner products> below by <positive-definite> <Hermitian forms>. A smooth rank-$r$ <vector bundle> is a smooth total space $E$ with a smooth projection $\pi:E\to M$, a real <vector space> structure on each <fiber> $E_p$, and an <open cover> with <vector bundle trivializations>
$$
\Phi_i:\pi^{-1}(U_i)\longrightarrow U_i\times\mathbb R^r.
$$
Each $\Phi_i$ is a <diffeomorphism> over $U_i$ and is linear on <fibers>. The overlap maps have the form $(p,v)\mapsto(p,A_{ji}(p)v)$, where $A_{ji}:U_i\cap U_j\to GL_r(\mathbb R)$ is smooth. The <rank of a vector bundle> may be specified separately on each <connected component>.
A smooth <fiber metric> is a <positive-definite> <symmetric bilinear form> $h_p$ on each <fiber>, with smoothly varying coefficients in every <local frame>. Equivalently it is a smooth section of $\operatorname{Sym}^2E^*$ which is positive definite at every point. Each trivialization supplies a local <fiber metric> $h_i$ by pulling back the Euclidean <inner product>.
A <smooth manifold> is Hausdorff and second-countable, hence <paracompact>. Choose a smooth, locally finite <partition of unity> $\{\phi_i\}$ subordinate to a trivializing cover, with $\phi_i\ge0$, $\operatorname{supp}\phi_i\subset U_i$, and $\sum_i\phi_i=1$. Define
$$
\boxed{h=\sum_i\phi_i h_i.}
$$
Extend each weighted term by zero outside its trivializing set; its support lies inside that set, so the extension is smooth. Local finiteness makes the sum smooth. At a point $p$ and for $0\ne v\in E_p$, every nonzero term $\phi_i(p)h_i(v,v)$ is positive and at least one such term occurs. Thus $h_p(v,v)>0$. This proves that \b[every smooth <vector bundle> admits a smooth <fiber metric>].
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