Proof of Lagrange's Mean Value Theorem? The Mean value theorem for Mapping says:

Let $f(x,y)$ be differentiable in $D$.  (D is open and connected).
For every $p=(x_1,y_1), q=(x_2,y_2)$ there exists a point $s \in [p,q]$ such that:
$$f(q)-f(p) = \nabla f(s)*(q-p)= f_x(s)(x_2-x_1)+f_y(s)(y_2-y_1)$$
Note: the interval $[p,q]$ is the following set of points $\{(1-t)p+tq |\ 0 \le t \le 1\}$

How may I prove that theorem? (it goes beyond my current level of studies)
Any help is appreciated :-)

Assuming$[p,q]\subseteq D$ (see @LurchedSawyer 's answer):
Let $g(t)=f((1-t)p+tq)$.
Then $$g'(t)=\nabla f((1-t)p+tq)*(q-p)$$ by the chain rule.  Using the usual 1 dimensional mean value theorem you may find a value $u\in [0,1]$ such that $$g'(u)=g(1)-g(0)=f(q)-f(p)$$
Setting $$s=(1-u)p+uq$$ gives:$$f(q)-f(p)=g'(u)=\nabla f(s)*(q-p)$$
