= Solution
A subset $U\subseteq\mathbb R^n$ is an <open set> when every $x\in U$ has some radius $r>0$ for which $B(x,r)\subseteq U$. If a <continuous function> takes only the values zero and one, continuity at $x$ with tolerance $1/2$ gives a neighborhood on which $|f(y)-f(x)|<1/2$. The two possible values differ by one, so $f(y)=f(x)$ throughout that neighborhood. Thus $f$ is <locally constant>, and the zero <linear map> is its <derivative>, since the difference $f(x+h)-f(x)$ is exactly zero for all sufficiently small $h$. Hence \b[$Df_x=0$ everywhere].
For the bound on $g$, parametrize the segment by $\gamma(t)=b+t(a-b)$, $0\le t\le1$. The <chain rule> gives
$$
(g\circ\gamma)'(t)=(Dg)_{\gamma(t)}(a-b),\qquad |(g\circ\gamma)'(t)|\le M\|a-b\|
$$
by the definition of <operator norm>. The <mean value theorem> applied to the real function $g\circ\gamma$ proves
$$
\boxed{|g(a)-g(b)|\le M\|a-b\|.}
$$
If the endpoints agree the estimate is immediate. No continuity of the <derivative> is required.
For the finite-line complement, each removed line $\ell_i$ is closed, so $V$ is open. Because $a,b\notin\ell_i$, there is a unique affine plane $P_i$ containing $a$ and $\ell_i$, and a unique affine plane $Q_i$ containing $b$ and $\ell_i$. Choose $c$ outside the union of these finitely many planes, using the permitted fact about finite plane covers. In particular $c\in V$. If $[a,c]$ met $\ell_i$ at a point $d$, then $d\ne a$, and the line through $a,d$ lies in $P_i$. Since $c$ is on that same line, it would belong to $P_i$, a contradiction. The same argument using $Q_i$ applies to $[c,b]$. Thus \b[$a$ and $b$ can be joined in $V$ by these two segments], proving <polygonal connectivity of a finite-line complement>.
Finally a continuous $f:V\to\{0,1\}$ has zero <derivative> by the first argument. Apply the proved segment estimate with $M=0$ separately on $[a,c]$ and $[c,b]$. It gives $f(a)=f(c)=f(b)$. Since $a,b$ were arbitrary, \b[$f$ is constant on $V$].
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