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 simplical 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][1].  Finally, $M_s$ determines $s$ by the uniqueness of the JSJ [decomposition][2].  As a remark - it is also possible to do this with hyperbolic manifolds (thus having trivial JSJ decomposition) again embedding in the three-sphere.  

  [1]: http://en.wikipedia.org/wiki/Antoine%2527s_necklace
  [2]: http://en.wikipedia.org/wiki/JSJ_decomposition