Corollary. Every \(m\)-manifold has topological dimension at most \(m\).
Solution
Let \(M\) be an \(m\)-manifold. It is Hausdorff and second-countable, hence locally compact and \(\sigma\)-compact.
Let \(K\subset M\) be compact. Cover \(K\) by finitely many coordinate neighborhoods and shrink them so that there are compact sets \[ K=K_1\cup\cdots\cup K_r \] with each \(K_i\) contained in one coordinate chart. Under that chart, \(K_i\) is homeomorphic to a compact, hence closed, subspace of \(\mathbb R^m\). Since \(\dim\mathbb R^m\le m\) and dimension does not increase on closed subspaces, \[ \dim K_i\le m. \] The finite closed-union theorem for dimension then gives \(\dim K\le m\).
Thus every compact subspace of \(M\) has dimension at most \(m\). Applying the \(\sigma\)-compact theorem E-MVB8P yields \[ \boxed{\dim M\le m}. \]