Stable Stiefel space is contractible

The case $n=0$ is the one-point space by [F1]. Suppose $n>0$. Let $O(e_r)=e_{2r-1}$ and $P(e_r)=e_{2r}$. For $Q\in \{O,P\}$ put $L_s^Q=(1-s)I+sQ$. Every $L_s^Q$ is injective. For $s>0$ and a nonzero finite vector whose largest nonzero coordinate is $r$, coordinate $2r$ of $L_s^{P}v$ is $s{v}_r\mathrm{\ne }0$. For $Q=O$ and $r>1$, coordinate $2r-1$ is $s{v}_r\mathrm{\ne }0$, while for $r=1$ coordinate $1$ is $v_1$. The case $s=0$ is immediate.

If $u=(u_1,\dots ,u_n)$ is a frame, injectivity of $L_s^Q$ makes $L_s^Q{u}_1,\dots ,L_s^Q{u}_n$ independent. Writing their column matrix as $A_s^Q$, the Gram-normalized matrix $A_s^Q((A_s^Q{)}^{\ast }{A}_s^Q{)}^{-1\mathrm{/}2}$ is an orthonormal frame and varies continuously in $(u,s)$ because positive-definite finite matrices have continuous inverse square roots. For each $Q\in \{O,P\}$ this homotopes $u$ to $Q(u)$, since $L_0^Q=I$ and $Q$ is an isometry.

Let $E=(e_2,e_4,\dots ,e_{2n})$. The vectors of $O(u)$ occupy odd coordinates and are orthogonal to the vectors of $E$. Therefore, for $0\le \theta \le \pi \mathrm{/}2$, the columns $\cos \theta O(u_i)+\sin \theta e_{2i}$ are orthonormal: their pairings are $({\cos }^2\theta +{\sin }^2\theta ){\delta }_{ij}$. They give a homotopy from $O(u)$ to the constant frame $E$.

Each finite Stiefel stage is compact Hausdorff because it is a closed subspace of a finite product of unit spheres, and its coordinate inclusion into the next stage is closed. Thus [F1] is a $k_{\omega }$-decomposition. By [F2], $V_n({\mathbb{F}}^{\mathrm{\infty }})\times I$ has the weak topology tested on $V_n({\mathbb{F}}^N)\times I$. On that product stage, the $O$ and $P$ homotopies in step 2.1 land respectively in $V_n({\mathbb{F}}^{2N-1})$ and $V_n({\mathbb{F}}^{2N})$, while step 3.1 lands in a finite stage containing also the first $2n$ coordinates. The compatible stagewise formulas are continuous, so [F2] proves ordinary continuity of both parity homotopies and of their concatenation from the identity to the constant frame $E$. This is the required contraction and uses no choice principle.