Every smooth manifold admits a riemannian metric

Apply the bundle-metric construction to $TM$; in the boundary case the boundary tangent theorem supplies its smooth rank-$n$ bundle. The needed chart selections can be made countably: form all nested relatively compact chart/frame tuples (half-balls at boundary), fix a countable basis, and use countable choice to select one eligible tuple over each basis member contained in such a chart. Their domains cover $M$.

In that countable cover, finite unions of compact closures exhaust $M$; choose least increasing indices putting each union in the interior of the next. Each compact annulus has a nonempty set of finite nested covering lists whose outer closures lie in a labelled trivialization and the adjacent open annular band. Countable choice selects these lists and then the corresponding compact-set bumps. The bands make the outer sets locally finite. Dividing the bump family by its positive smooth sum gives weights ${\rho }_i$ summing to one with closed supports inside their labelled charts. The boundary partition construction uses the same argument with restricted half-space bumps. Thus the bundle-metric supplier’s selections need only the assumed countable choice.

On chart $i$ take the Euclidean frame metric $g_i$ and extend ${\rho }_i{g}_i$ by zero; this is smooth because its support is closed inside the chart. The locally finite sum $g={\sum }_i{\rho }_i{g}_i$ is smooth and symmetric. For $v\mathrm{\ne }0$ at $p$, each term is nonnegative and some ${\rho }_i(p)>0$, whence $g_p(v,v)\ge {\rho }_i(p)(g_i{)}_p(v,v)>0$. Thus $g$ is Riemannian. Empty $M$ uses the empty metric; rank zero uses the zero fibre form.