My attempt to solve Problem 4-3 (Lee's Introduction to Topological Manifolds, $1$st edition) "Show that any $n-$manifold is a disjoint union of countably many connected $n-$manifolds" is the following:
Let $M$ be a $n-$dimensional manifold and $p_1\in M$. There exists a(n) (open) neighborhood around $p_1$, let's say $U_1$. Define a relation in $M$ such that if $V$ is open subset of $M$, then
$$V \sim U_1 \Leftrightarrow \exists \; W_1,\ldots,W_k\subsetneqq M \; open/locally \; Euclidean: U_1\cap W_1,W_1\cap W_2,\ldots,W_k\cap V\neq\emptyset$$
This is an equivalence relation, therefore it provides a partition of M.
I suppose that each equivalence class is a connected component or, in this case, a connected n-manifold: $U_1$ is connected because it is homeomorphic to $\mathbb{R}^n$ (after some definitional modifications), $U_1\cap W_1$ is connected for the same reason, so $W_1$ must be connected too, and by induction V is connected.
If $M\backslash[U_1]\neq\emptyset$, then there exists a point $p_2\in M\backslash[U_1]$ with a(n) (open) neighborhood $U_2$ homeomorphic to $\mathbb{R}^n$. So, $[U_2]$ would be the second connected component/$n-$manifold.
By continuing this procedure, $M$ is a disjoint union of connected $n-$manifolds. The countability derives from second-countability of $M$.
Is my suggested solution correct?
Best Answer
Claim 1: $M$ has only countably many components.
Proof: If $M$ had uncountably many components $(M_{\alpha})_{\alpha\in A},$ then each $M_{\alpha}$ is open and disjoint. Hence, as you said, $M$ cannot be second countable.
Claim 2: Each component of $M$ is an $n$-manifold.
I'm assuming here that Lee has it as a definition that the dimension of any two charts is always the same $n$ (otherwise, there's no reason why, say, $S^1\coprod S^2$ shouldn't be a manifold).
Let $M_k$ be a component of $M$and fix $x\in M_k$. Since $M$ is a manifold, there is $x\in U\subseteq M$ open and a homeomorphism $\varphi:\mathbb{R}^n\to U$. Then, by continuity, $U$ is connected and by assumption, it contains $x$. Hence, $U\subseteq M_k$. Thus, each $M_k$ is locally Euclidean.
To see that $M_k$ is second countable, let $(U_j)_{j\in \mathbb{N}}$ be a basis for the topology of $M$ and note that $(M_k\cap U_j)_{j\in \mathbb{N}}$ must be a basis for the subspace topology on $M_k$.