The rational homology of a closed smooth manifold is finite-dimensional in each degree

(A finite cellular model.) By [F1] there is a homotopy equivalence $M\to X$ onto a finite CW complex, inducing isomorphisms $H_k(M;G)\cong H_k(X;G)$ for every abelian coefficient group $G$ [F1, F3]. The cellular chain group $C_k(X)$ in degree $k$ is free abelian on the finitely many $k$-cells, hence finitely generated and free [F2].

(Finite generation over $\mathbb{Z}$.) Let $Z_k\subseteq C_k(X)$ be the group of cellular $k$-cycles and $B_k\subseteq C_k(X)$ the group of cellular $k$-boundaries, where $B_k$ is the image of $C_{k+1}(X)$ under the boundary map. Since $C_k(X)$ is finitely generated free, its subgroup $Z_k$ is free and finitely generated by [F4]; since $C_{k+1}(X)$ is finitely generated, its image $B_k$ is finitely generated as well. Hence $H_k(X;\mathbb{Z})=Z_k\mathrm{/}{B}_k$ is a quotient of a finitely generated abelian group and is finitely generated, and by step 1.1 the same holds for $H_k(M;\mathbb{Z})$ [F1, F2, F4].

(Finite dimension over $\mathbb{Q}$.) Compute cellular homology directly with coefficients $\mathbb{Q}$: each cellular chain group is a finite-dimensional vector space on the finitely many cells by F2 and F3. Its kernel is a subspace and the homology is the quotient of that kernel by the image of the next boundary, so it is finite dimensional. Step 1.1 transfers this conclusion to $H_k(M;\mathbb{Q})$. Every subspace of a finite-dimensional vector space is finite dimensional; in particular this applies to the span of countably many classes and to degree two. No identification of integral cycle groups after tensoring, and hence no unstated flatness assertion, is needed.

Therefore $H_k(M;\mathbb{Z})$ is finitely generated and $H_k(M;\mathbb{Q})$ is finite-dimensional for every $k$, with the stated consequences for $H_2$ and for subspaces spanned by countably many classes.