Counting connected manifolds Apparently, there is the following fact:

The set of homeomorphism classes of connected manifolds has the same cardinality $c$ as that of $\mathbb R$.

I find it to be interesting; but would be happier to see a proof, and would be grateful for a reference somewhere.
 A: There are countably many compact topological manifolds, as was shown by Cheeger-Kister in Counting topological manifolds, Topology 9 (1970) 149–151, https://doi.org/10.1016/0040-9383(70)90036-4. The proof depends on results of Edwards-Kirby of deformation of homeomorphisms. A sketch of the proof is here.
Incidentally, a good discussion of counting non-metrizable manifolds is in section 3 of Gauld's paper Manifolds at and beyond the limit of metrisability section 3.
A: I am worried that this may not be true in TOP (the topological category).  Let us work in PL (piece-wise linear) instead.  Every PL manifold can be expressed as a locally finite simplicial complex, with at most a countable number of vertices.  Thus there are at most $2^N$ of these.  (Here $N$ is the natural numbers.)  I think I will leave the lower bound as an exercise -- I have a way of doing this by encoding binary sequences using generalized Heegaard splittings of three-manifolds, but that is a hack.  I am sure that there is a more beautiful way to give the lower bound just using non-compact surfaces.
I looked, but could not find a reference.
EDIT: Just in case the above is too brief here is an "easier" exercise. The number of simple locally finite graphs on at most a countable number of vertices is again $2^N$.  [Hint: consider the adjacency matrix.] [In fact, the locally finite hypothesis is not necessary in this case.]
EDIT2: After thinking a bit more about my three-manifold examples, I realized the statement can be simplified a bit.  You can embedd the set of binary sequences into the set of homeomorphism classes of submanifolds of $S^3$ as follows: Choose your two favorite distinct hyperbolic links in $S^3$, each with two components.  Lets call these links $L_0$ and $L_1$.  Given a binary sequence $s \colon N \to \{0,1\}$ form a three manifold $M_s$ by gluing copies of $L_0$ and $L_1$ as instructed by $s$.  Note that $M_s$ embeds in the three-sphere; the complement looks a bit like Antoine's necklace.  Finally, $M_s$ determines $s$ by the uniqueness of the JSJ decomposition.  As a remark - it is also possible to do this with hyperbolic manifolds (thus having trivial JSJ decomposition) again embedding in the three-sphere.
Ah - following the link that algori provides reminds me that there are (up to homeomorphism) uncountably many Whitehead manifolds - open, contractable submanifolds of $S^3$.  The point of all of these examples is that it is "easy" to encode information in the end of a three-manifold.  I'll leave my examples here, as they are a useful warm-up to understanding the Whitehead manifolds.
A: Upper bound (assuming the manifolds are second countable): every manifold admits a complete metric, and the "set" of isometry classes of complete separable metric spaces is of cardinality continuum. Indeed, every such a space is a completion of a countable metric space, and there is only continuum of metrics on a countable set.
Lower bound: there is a continuum of non-isomorphic fundamental groups of manifolds. E.g. any countable sum of cyclic groups is a first homology group of a connected manifold: just take a connected sum of an appropriate countable collection of lens spaces.
