Is it true that all homotopy groups of a manifold are countable? It is well-known that the first homotopy group (fundamental group) of a manifold is countable. I would like to know this for higher homotopy groups of a manifold. i.e.

Question: Is it true that all homotopy groups of a manifold are countable?

I think one can do same process in the proof of countability of fundamental group for higher homotopy groups. i.e. using Lebesgue Number Lemma. See SM J. M. Lee. Am I right? If not any proof or reference for the proof.

Every second-countable ANR has has the homotopy type of a countable CW complex. Thus every second-countable (i.e. metrisable) manifold, being an ANR, has the homotopy type of a countable CW complex.
Let $M$ be a second-countable manifold with a chosen basepoint. Note that $M$ is therefore separable and metrisable. Let $C_*(S^n,M)$ be the set of pointed maps $S^n\rightarrow M$. If $C_*(S^n,M)$ is given the uniform topology, then it becomes a separable metric space, and in particular is second-countable.
On the other hand, $C_*(S^n,M)$ in the compact-open topology is an ANR, and so homotopy equivalent to a CW complex. But since $S^n$ is compact, the compact-open and uniform topologies on $C_*(S^n,M)$ coincide. Therefore by the above $C_*(S^n,M)$ is homotopy equivalent to a countable CW complex. Thus it has countably many path-components, each of which is open. In particular
$$\pi_0(C_*(S^n,M))\cong \pi_n(M)$$
is countable.
