$\newcommand{\Z}{\mathbf{Z}}$The Chern classes may be viewed as a map $c: \mathrm{BU}(n) \to \prod_{i=1}^n K(\Z,2i)$. Call the target $X_n$. Complex conjugation defines a $\Z/2$-action on $\mathrm{BU}(n)$, whose fixed points are $\mathrm{BO}(n)$. Give $X_n$ the $\Z/2$-action defined by viewing $K(\Z, 2n)$ as $\Omega^\infty(S^{2n,n} \mathrm{H}\Z)$$\Omega^\infty(\Sigma^{2n,n} \mathrm{H}\Z)$, where $S^{2n,n} = S^{n + n\sigma}$ (here, $S^\sigma$ is the one-point compactification of the sign representation of $\Z/2$), and $\mathrm{H}\Z$ is given the sign actionEilenberg-Maclane spectrum associated to the constant Mackey functor. I thinkforgot to write this in the previous version of this answer, but (as in Lennart's answer) there is a projection map $\Phi^{C_2} \mathrm{H}\Z\to \mathrm{H}\mathbf{F}_2$. The map $c$ is then $\Z/2$-equivariant, and taking $\Z/2$-fixed points and composing with the above projection yields the map $\mathrm{BO}(n) \to \prod_{i=1}^n K(\Z/2, i)$ given by the Stiefel-Whitney classes. I haven't done this explicitly, but to check that these are in fact the Stiefel-Whitney classes, one reduces to the case of line bundles (by the splitting principle); then, it follows from the fact that $\mathrm{BU}(1)$ is equivariantly equivalent to $\Omega^\infty \Sigma^{2,1} \mathrm{H}\Z$ (and this identification is given by the first Chern class), and taking fixed points and projecting gives the identification of $\mathrm{BO}(1)$ with $\Omega^\infty \Sigma \mathrm{H}\mathbf{F}_2$ (where this identification is given by the first Stiefel-Whitney class). [This is essentially the same as Lennart's answer.]
