= Solution
Put $c=e^{-\gamma}$, where $\gamma$ is the <Euler--Mascheroni constant>, and let $f(u)=w(u)-c$. The <Buchstab function> equation becomes
$$
\frac{d}{du}(u f(u))=f(u-1)\qquad(u>2).
$$
The given <Gamma function> bound implies $u f(u)\to0$. Indeed,
$$
\Gamma(u+1)=\int_0^\infty e^{-x}x^u\,dx\ge\int_u^{u+1}e^{-x}x^u\,dx\ge e^{-u-1}u^u,
$$
so $u/\Gamma(u+1)\to0$.
Suppose first that $f\ge0$ throughout an interval $[u,u+1]$, with $u\ge1$. The <delay differential equation> and integration successively on $[u+1,u+2]$, $[u+2,u+3]$, and so on propagate nonnegativity to the entire tail $[u,\infty)$. In particular, $v f(v)$ is nonnegative and nondecreasing for $v\ge u+1$. Its limit is zero, so it is identically zero there. Thus $f=0$ on that tail.
The <delay differential equation> then propagates this equality backwards: if $f=0$ for $v\ge a$, then $f(v-1)=0$ for $v>\max(a,2)$. Repeating finitely many times would give $f=0$ on $(1,2)$, contradicting $f(v)=1/v-c$ there. Hence every interval $[u,u+1]$ contains a point where $f<0$.
If instead $f\le0$ on $[u,u+1]$, the same propagation makes $v f(v)$ nonpositive and nonincreasing on the tail. Its limit zero again forces it to vanish, giving the same contradiction. Thus every such interval also contains a point where $f>0$. \b[The oscillation of the Buchstab function is therefore strict on every unit interval]:
$$
\boxed{\exists u^+,u^-\in[u,u+1]:\quad w(u^+)>e^{-\gamma},\quad w(u^-)<e^{-\gamma}.}
$$
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