A proof could go as follows:
Let $\Omega \in\mathbb{R^3}$, and consider the force field (which is a vector field) $\textbf F:\Omega \rightarrow \mathbb{R^3} $. Now, consider a particle $P$, of mass $m$, and let $\gamma:[t_1,t_2]\rightarrow \Omega$, with $\gamma \in C^2([t_1,t_2])$ be the curve representing the path that $P$ follows in space. 418
We assume that $\textbf F$ is the total force acting on P
(Remember that the symbol $\cdot$ indicates the scalar product)
The work done by $F$ on $P$ between the instants $t_1,t_2$, is defined as the line integral of $F$ over $\gamma$; that is,
$$W=\int_\gamma \textbf F=\int_{t_1}^{t_2} \textbf F(\gamma(t))\cdot\gamma'(t)dt $$
By the second law, $\textbf F(\gamma(t))=m\textbf a(t)=m\gamma''(t)$, thus, if $\textbf v(t)=\gamma'(t)$ is the velocity of $P$, $$ \int_{t_1}^{t_2} \textbf F(\gamma(t))\cdot\gamma'(t)dt=\int_{t_1}^{t_2}m\gamma''(t)\cdot \gamma'(t)=\int_{t_1}^{t_2}m\textbf v'(t)\cdot \textbf v(t) $$
It's clear that $\textbf v(t)$ is a vector, hence we'll write it in terms of its coordinates: $\textbf v(t)=(v_x(t),v_y(t),v_z(t))$, and of course its derivative, the acceleration, can be written as $\textbf v'(t)=(v'_x(t),v'_y(t),v'_z(t))$ . Hence, the scalar product between the velocity and the acceleration is $v_x(t)v'_x(t)+v_y(t)v'_y(t)+v_z(t)v'_z(t)$. Thus,
$$\int_{t_1}^{t_2}m\textbf v'(t)\cdot \textbf v(t)=m\int_{t_1}^{t_2}(v'x(t),v'_y(t),v'_z(t))\cdot (v_x(t),v_y(t),v_z(t))=m\int{t_1}^{t_2}(v_x(t)v'x(t)+v_y(t)v'_y(t)+v_z(t)v'_z(t))dt=m\int{t_1}^{t_2} v_x(t)v'x(t)dt+ m\int{t_1}^{t_2} v_y(t)v'y(t)dt+m\int{t_1}^{t_2}v_z(t)v'_z(t)dt $$.
Now it's easy to see (if not, just integrate by parts) that $\int_{t_1}^{t_2} v_x(t)v'_x(t)dt=\frac{1}{2}(v_x^2(t_2)-v_x^2(t_1))$, and the same for the other two integrals. 804