I believe each of these arguments will work.

**Argument 1:** Consider $S^n \subset \Bbb R^{n+1} \subset S^{n+1}$, where the last inclusion is
given by the upper hemisphere (which is homeomorphic to $\Bbb R^{n+1}$. This gives rise to a string of inclusions

$$
\cdots \subset F(S^n,k) \subset F( \Bbb R^{n+1},k) \subset F(S^{n+1};k) \subset F(\Bbb R^{n+2},k)\subset \cdots
$$
The colimit this system is identified with both $F(S^\infty,k)$ and $F(\Bbb R^\infty,k)$.

**Argument 2:** The projection onto the last coordinate defines a Hurewicz fibration
$$
F(S^\infty,k) \to S^\infty
$$
whose fiber is identified with $F(\Bbb R^\infty,k)$. But $S^\infty$ is contractible, which shows that the fibration is trivializable. In particular, one has a homotopy equivalence
$$
F(S^\infty;k) \simeq F(\Bbb R^\infty,k)\times S^\infty\, .
$$
Now use the fact that $F(\Bbb R^\infty,k)$ is contractible.