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\begin{document}
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\noindent{\bf Rudin, page 239/9:}
Fix any point $p\in E$.  Let 
$$
K = \{ x\in E: f(x) = f(p)\}.
$$
I will show that $K = E$, which implies of course that $f$ is constant.
Now $p\in K$, so $K\neq \emptyset$.  Moreover, $f$ is continuous because
it is differentiable, so $K = f^{-1}(f(p))$ is closed (i.e. the inverse
image of a closed set by a continuous function is closed).  In particular,
$E - K$ is open.  If I can show that $K$ is also open, then 
$$
E = K \cup (K-E)
$$
will express $E$ as a disjoint union of two open sets.  But $E$ is connected
by hypothesis, so it will follow that $K-E = \empty$ and therefore $E=K$.

So to summarize, it's enough to show that $K$ is open.  Let $x\in K$ be
any point. Because $E$ itself is open, we can choose $r>0$ such that
$N_x(r) \subset E$.  Let $y\in N_x(r)$ be any other point and 
$$
h:[0,1]\to \R
$$
be given by $h(t) =f(t x + (1-t) y)$.  Then $h(1) = f(x) = f(p)$ and
$h(0) = f(y)$.  Moreover, as the composition of two differentiable functions
$h$ itself is differentiable, with
$$
h'(t) = Df(tx + (1-t)y) \cdot \frac{d}{dt}(tx + (1-t)y)
= \mathbf0\cdot (x-y) = 0
$$
for all $t\in [0,1]$.  It follows (from one variable calculus) that
$h$ is constant.  In particular $f(y) = h(1) = h(0) = f(p)$.  And $y\in
N_r(x)$ was arbitrary, so we conclude that $N_r(x)\subset K$.  As $x$ was
arbitrary, too, it follows that $K$ is open.

\end{document}