If $X$ and $Y$ are homotopy equivalent, then are $X \times \mathbb{R}^{\infty}$ and $Y \times \mathbb{R}^{\infty}$ homeomorphic? Let $X$ and $Y$ be reasonable spaces.  Since $\mathbb{R}^{\infty}$ is contractible, 
$$
X \times \mathbb{R}^{\infty} \cong Y \times \mathbb{R}^{\infty} \;\;\; \implies \;\;\; X \simeq Y.
$$
Is the converse also true?
My vague intuition: the factors of $\mathbb{R}^{\infty}$ provide so much extra room that there will never be a geometric obstruction to producing a homeomorphism.  Evidently, there is no homotopy-theoretic obstruction, so maybe the converse is true.
On the other hand, I really have no idea and could be missing something basic.  For example, the plane with two punctures is homotopy equivalent to a wedge of two circles.  However, I do not know about a homeomorphism
$$
(\mathbb{C} - \{0, 1\}) \times \mathbb{R}^{\infty} \overset{?}{\cong} (S^1 \vee S^1) \times \mathbb{R}^{\infty}.
$$

Clarification about the meaning of $\mathbb{R}^{\infty}$ and the intent of the question
When I wrote the question, I had in mind the infinite union $\cup_n \mathbb{R}^n$ inside the product $\mathbb{R}^{\mathbb{N}}$.  However, since I would like the answer to the question to be "yes," I am also interested in other versions of $\mathbb{R}^{\infty}$.
I asked the question because of an algebraic limiting construction in a paper I'm writing, and I felt that a topological version of the limit would be satisfying.  The algebraic version is already working for spaces like $X \times \mathbb{C}^n$ for large-enough $n$, and converges algebraically to some limiting group, but this doesn't give too many hints about the topology I should use on $\mathbb{C}^{\infty}$, or if the limit can even be considered topologically.
My application involves singular homology, and the direct limit topology is well-suited to this application, but other choices may be as well.
 A: This question is answered by two classical theorems of infinite-dimensional topology, which can be found in the books of Bessaga and Pelczynski, Chigogidze or Sakai.
Factor Theorem. For any Polish absolute neighborhood retract $X$ (= neighborhood retract of $\mathbb R^\omega$) the product $X\times\mathbb R^\omega$ is an $\ell_2$-manifold.
Classification Theorem. Two $\ell_2$-manifolds are homeomorphic if and only if they are homotopy equivalent.
So, the reasonable space in your question should read as a Polish absolute neighborhood retract. By the way, this class of spaces includes all (countable locally) finite simplicial complexes, mentioned in the answer of Igor Rivin.

Remark on possible generalizations. Chapter IX of the book of Bessaga and Pelczynski contains generalizations of the above two theorems to manifolds modeled on normed spaces $E$, which are homeomorphic to $E^\omega$ or $E^{\omega}_0:=\{(x_n)_{n\in\omega}\in E^\omega:\exists n\in\omega\;\forall k\ge n\;(x_k=0)\}$.
Chapter 7 of Chigogidze's book contains generalizations of the Factor and Classification Theorems to manifolds modeled on uncountable products of lines.
Chapter 5 of Sakai's book contains generalizations of the Factor and Classification Theorems to manifolds modeled on the direct limits $\mathbb R^\infty$ and $Q^\infty$ of Euclidean spaces and Hilbert cubes, respectively.
So, typical generalizations of Factor and Classification Theorems look as follows:

Theorem. Let $E$ be a reasonable model space (usually it is an infinite-dimensional locally convex space with some additional properties).
$\bullet$ For any neighborhood retract $X$ of $E$ the product $X\times E$ is an $E$-manifold.
$\bullet$ Two $E$-manifolds are homeomorphic if and only if they are homotopically equivalent.


Depending on the meaning of your model space $\mathbb R^\infty$ the meaning of a reasonable space also changes. If $\mathbb R^\infty$ is the direct limit of Euclidean spaces, then a reasonable space means an absolute neighborhood extensors which is a direct limits of finite dimensional compacta. If $\mathbb R^\infty$ is the union of the Euclidean spaces in the countable product of lines, then a reasonable space is a locally contractible space which is the countable union of finite-dimensional compact subsets.
A: I believe the answer is YES, where reasonable =  finite simplicial complex, based on the work of Chapman on Hilbert Cube manifolds. Namely, I believe that it is a 1973 definition of Chapman that a Hilbert cube manifold of "type $Q$" if it is homeomorphic to $Q \backslash A,$ where $Q$ is the Hilbert cube $[0, 1]^\infty,$ and $A$ is a closed subset of $W = \{x\in Q \left| x_1 = 1\right.\}$ Now, it seems to be a theorem (of Chapman, 1972) that a manifold is of type $Q$ iff it is of the form $|Y| \times Q,$ where $Y$ is a locally finite simplicial complex, and further, two such type $Q$ manifolds are homeomorphic if and only if they are properly homotopic (clearly a homotopy induced by the homotopy of $|Y|$ is proper).
Chapman, T.A., Contractible Hilbert cube manifolds, Proc. Am. Math. Soc. 35, 254-258 (1972). ZBL0259.58004.
Chapman, T.A., On the structure of Hilbert cube manifolds, Compos. Math. 24, 329-353 (1972). ZBL0246.57005.
A: This (Brown--Cohen, A proof that simple-homotopy equivalent polyhedra are stably homeomorphic) gives a partial answer (maybe subsumed by the answer by Igor Rivin).
