The proof is from Folland's Advanced Calculus. The theorem states: Let $f$ be a function defined on an open set in $\mathbb{R}^n$ that contains the point $\vec{a}$. Suppose the partial derivatives $\partial_jf(\vec{a})$ is differentiable at $\vec{a}$.
Proof:
For the sake of simplicity, we let $n=2$. We want to show that $\dfrac{f(\vec{a}+\vec{h})-f(\vec{a})-\vec{c}\cdot \vec{h}}{|\vec{h}|} \to 0$ as $h \to 0$ where $c=(\partial_1f(\vec{a}),\partial_2f(\vec{a}))$.
We analyze the increment $f(a+h)-f(a)$ by making the change one variable at a time: $f(\vec{a}+\vec{h})-f(\vec{a})=[f(a_1+h_1,a_2+h_2)-f(a_1,a_2+h_2)]-[f(a_1,a_2+h_2)-f(a_1,a_2)]$.
We assume that $\vec{h}$ is small enough so that the partial derivatives $\partial_jf(\vec{x})$ exist whenever $|x-a|\leq |h|$. In this case, we can use the one-variable mean value theorem to express what is in the brackets in terms of partial derivatives of $f$. If we let $g(t)=f(t,a_2+h_2)$, then $f(a_1+h_1,a_2+h_2)-f(a_1,a_2+h_2)=g(a_1+h_1)-g(a_1)=g'(a_1+c_1)h_1=\partial_1f(a_1+c_1,a_2+h_2)h_2$. Similarly, $f(a_1,a_2+h_2)-f(a_1,a_2)=\partial_2f(a_1,a_2+c_2)$ for some number $c_1$ lying between $0$ and $h_1$. Therefore,
$\dfrac{f(\vec{a}+\vec{h})-f(\vec{a})-\vec{c}\cdot \vec{h}}{|\vec{h}|} =[\partial_1f(a_1+c_1,a_2+h_2)-\partial_1f(a_1,a_2)]\dfrac{h_1}{|\vec{h}|}+[\partial_2f(a_1,a_2+c_2)-\partial_1f(a_1,a_2)]\dfrac{h_2}{|\vec{h}|}$
Now if we let $h \to 0$, the expressions in the bracket tend to $0$ since the partial derivatives are continuous at $vec{a}$ and the ratios $\dfrac{h_1}{|\vec{h}|}$ and $\dfrac{h_2}{|\vec{h}|}$ are bounded by $1$ in absolute value. Therefore f is differentiable at $\vec{a}$. $\square$
My questions are in the last statement. How does knowing that the partial derivatives are continuous imply they will go to $0$ when $h\to 0$? Also, how do we know that the ratios $\dfrac{h_1}{|\vec{h}|}$ and $\dfrac{h_2}{|\vec{h}|}$ are bounded by $1$?
First of all, note that
$$|\vec{h}|=\sqrt{h_1^2+h_2^2}\ge |h_1|.$$ Thus
$$\frac{|h_1|}{|\vec{h}|}\le 1.$$
On the other hand
$$\partial_1f(a_1+c_1,a_2+h_2)-\partial_1f(a_1,a_2)\to 0$$ as $h\to 0$ because of continuity. Note that $c_1$ lies between $0$ and $h_1.$ Thus, the above quantity goes to zero can be written simply as
$$\lim_{(h_1,h_2)\to (0,0)} \partial_1f(a_1+c_1,a_2+h_2)=\partial_1f(a_1,a_2),$$ what is the definition of continuity at the point $(a_1,a_2).$