Good question! (+1 for the same)
Wikipedia is correct here. And it will be best to explain this using an actual concrete example. Let $f$ be differentiable at $a$ and let $f'(a) > 0$. Using limit definition of derivative and taking $\epsilon = f'(a)/2$ we see that there is a $\delta > 0$ such that $$\left|\frac{f(a + h) - f(a)}{h} - f'(a)\right| < \frac{f'(a)}{2}$$ for all $h$ with $0 < |h| < \delta$. This means that $$\frac{f(a + h) - f(a)}{h} > \frac{f'(a)}{2} > 0$$ for all $h$ with $0 < |h| < \delta$. Therefore $f(a + h) - f(a)$ has same sign as that of $h$ for $0 < |h| < \delta$. Thus if $h > 0$ then $f(a + h) > f(a)$ and if $h < 0$ then $f(a + h) < f(a)$.
Therefore there is a neighborhood $I = (a - \delta, a + \delta)$ of $a$ such that if $x \in I, x > a$ then $f(x) > f(a)$ and if $x \in I, x < a$ then $f(x) < f(a)$. If a function $f$ has such a behavior in some neighborhood of $a$ then we say that $f$ is strictly increasing at point $a$. The property of strictly increasing at a point is a local property and we have proved that:
If $f'(a) > 0$ then $f$ is strictly increasing at $a$.
Contrast this definition of local property of increasing at point with the global property of increasing on an interval. A function $f$ defined on an interval $I$ is said to be strictly increasing on $I$ if $x, y \in I, x < y$ implies $f(x) < f(y)$. Now it appears obvious that if a function $f$ is strictly increasing at every point of an interval $I$ then it must be strictly increasing on $I$. This fact is true but the proof is not trivial and requires us to use completeness property of real numbers.
However if we assume $f$ to be differentiable on $I$ then we can prove the following theorem:
If $f$ is differentiable on $I$ and $f'(x) > 0$ for all $x \in I$ then $f$ is strictly increasing on $I$.
Proof is immediate from mean value theorem because $$f(x) - f(y) = f'(c)(x - y)$$ so that $f(x) - f(y)$ has same sign as that of $x - y$. Note that a positive derivative at each point implies that $f$ is strictly increasing at each point and using mean value theorem we have proved that $f$ is strictly increasing on $I$.
The above theorem is thus an example of deducing global property from a local one. Also note that the above theorem is a special case of the general theorem we discussed earlier which says that a function which is strictly increasing at each point of an interval is strictly increasing on that interval irrespective of the fact whether the function is differentiable or not.
The above demonstration also shows that mean value theorem is a very powerful one and should not be considered obvious or trivial. And like most non-trivial theorems of calculus its proof also requires the completeness property of real numbers.
