Set of Increasing Functions on an Interval is Closed To prove the existence of nowhere monotone continuous functions, one step is to show that for any interval $I \subset [0,1]$ of positive length, the set $A_I= \{f \in C([0,1]): f|_I$ is increasing$\}$ is closed (i.e. see http://www.apronus.com/math/nomonotonic.htm).  I am not sure how to do prove this.
My idea is that if $(f_n) \in A_I$ converges to $f$, then for all $x <y \in I$, consider $f(x)-f(y) = (f(x)-f_n(x))+(f_n(x)-f_n(y))+(f_n(y)-f(y))$.  The first and third brackets can be made arbitrarily small and the middle bracket is negative, so the left side intuitively should also be negative, which would show that $f \in A_I$.  I am not sure how to actually prove this.
Thank you.

Your approach perhaps isn’t quite the simplest one, but it can be made to work:
$$\begin{align*}
f(x)-f(y) &= \big(f(x)-f_n(x)\big)+\big(f_n(x)-f_n(y)\big)+\big(f_n(y)-f(y)\big)\\
&\le\big(f(x)-f_n(x)\big)+\big(f_n(y)-f(y)\big)
\end{align*}$$
for all $n\in\Bbb N$, so
$$\begin{align*}
f(x)-f(y)&\le\lim_{n\to\infty}\Big(\big(f(x)-f_n(x)\big)+\big(f_n(y)-f(y)\big)\Big)\\
&=\lim_{n\to\infty}\big(f(x)-f_n(x)\big)+\lim_{n\to\infty}\big(f_n(y)-f(y)\big)\\
&=0+0\\
&=0\;,
\end{align*}$$
which is exactly what you want.
