Let $[a,b]$ be any closed interval and $f$ satisfying the mean value property for every open interval $(a',b')$ contained within $[a,b]$. Let $a_n = a - \frac{1}{n}$ and $b_n = b - \frac{1}{n}$ for large values of $n$. Then we have that for each $n$, there's a point $c_n$ in $(a_n, b_n)$ so that $f'(c_n) = \frac{f(b_n) - f(a_n)}{b_n - a_n}$. It's easy to see by continuity of $f$ on $[a,b]$ that the right hand side converges to $\frac{f(b) - f(a)}{b-a}$. I guess the question then becomes:
1) Is the sequence $\{c_n\}$ convergent and,
2) Even if it is, do we have $\lim_{n\to\infty} f'(c_n) = f'(c)$.
An imposition of continuity on $f'$ would be enough for 2). Now for $1$ we argue as follows: $\{c_n\}_n$ is a sequence in $[a,b]$ and so it has a convergent subsequence by compactness. Then this new sequence, say $\{c_{n_k}\}$ satisfies
$\lim_{n_k \to \infty} f'(c_{n_k}) = \frac{f(b) - f(a)}{b-a}$, and $\lim_{n_k} c_{n_k} = c$, for some point $c \in [a,b]$. By continuity, we have that there's a point $c \in [a,b]$ so that $f'(c) = \frac{f(b) - f(a)}{b-a}$, which isn't quite the mean value theorem.