Let $\mathbb{R}^\infty$ be the union $\cup \mathbb{R}^n$(with inclusions). Let $F(\mathbb{R}^\infty,k)$ denote the $k$-th configuration space of $\mathbb{R}^\infty$, i.e., the subspace of $(\mathbb{R}^\infty)^k$ consisting of all $k$-tuples of distinct points of $\mathbb{R}^\infty$. Let permutation group $\Sigma_k$ act on $F(\mathbb{R}^\infty,k)$. How to obtain $F(\mathbb{R}^\infty,k)/\Sigma_k\cong B\Sigma_k$, the classifying space of $\Sigma_k$?
Is $F(\mathbb{R}^\infty,k)$ contractible? Is an arbitrary configuration space $F(X,k)$ contractible?
This may not be the easiest approach, but I'll employ one of the only tools I know for handling configuration spaces of manifolds - The Fadell-Neuwirth fibration. This is really just a rehashing of the usual proof that the configuration spaces of a genus $g\geq 1$ surface, or the plane, is an Eilenberg Maclane space. Hopefully I haven't made a mistake...
I will take as given that $\mathbb{R}^{\infty}\setminus \mathbf{Q}_m$ is a contractible space, where $\mathbf{Q}_m$ is a set of $m$ distinct points in $\mathbb{R}^{\infty}$. This is the case, I believe, because $\mathbb{R}^{\infty}\setminus \mathbf{Q}_m$ has the homotopy type of an $m$-fold wedge product of infinite dimensional spheres $S^{\infty}$, and the wedge product of a collection of contractible spaces is itself contractible under some fairly weak conditions (certainly the fact that infinite-dimensional sphere are manifolds is sufficient).
Let $M$ be a manifold and let $F_{m,n}M = F_{0,n}(M\setminus\mathbf{Q}_m)$ where $F_{0,n}X$ is the configuration space of $n$ points in the space $X$, $$F_{0,n}X = X^n\setminus\Delta$$
and the subspace $\Delta$ here is the fat diagonal $\{(x_1,\ldots, x_n)\mid \exists i\neq j,\:\: x_i=x_j\}$, the subspace of points whose coordinates coincide for at least one pair.
Fadell and Neuwirth showed in their 1962 paper [1] that for a manifold $M$ and natural numbers $m,n,r$ with $1\leq r < n$, there exists a locally trivial fibration $p\colon F_{m,n}M\to F_{m,r}M$ with fiber homeomorphic to $F_{m+r,n-r}M$.
Letting $r=1$, and recalling that $F_{m,1} M = M\setminus \mathbf{Q}_m$, this gives us the long exact sequence in homotopy $$\cdots \to \pi_{k+1} M\setminus\mathbf{Q}_m \to \pi_k F_{m+1,n-1}M \to \pi_k F_{m,n}M\stackrel{p_*}{\to} \pi_k M\setminus\mathbf{Q_m} \to \cdots$$ where we see that if $\pi_{k+1} M\setminus\mathbf{Q}_m\cong\pi_k M\setminus\mathbf{Q}_m\cong 0$ then there is an induced isomorphism $$\pi_k F_{m+1,n-1}M \stackrel{\cong}{\to} \pi_k F_{m,n}M.$$
If this holds for all $m\geq 0$ then we get a chain of isomorphism $$\pi_k F_{0,n}M \cong \pi_k F_{1,n-1}M \cong \cdots \cong \pi_k F_{n-1,1} M$$ where we have already established that $\pi_k F_{n-1,1}M=\pi_k M\setminus\mathbf{Q}_{n-1}=0$ for all $k\geq 1$.
All of the above holds for $M=\mathbb{R}^{\infty}$ and so we can conclude that every homotopy group of $F_{0,n}\mathbb{R}^{\infty}$ is trivial. As $F_{0,n}\mathbb{R}^{\infty}$ is homotopy equivalent to a CW complex, we can conclude via Whitehead's theorem that $F_{0,n}\mathbb{R}^{\infty}$ is contractible.
[1] E. Fadell and L. Neuwirth (1962), Configuration spaces. Math. Scand 10, 111 - 118