Let $f(x,y):\mathbb{R^2} \rightarrow \mathbb{R}$, where $A \subset \mathbb{R^2}$ is convex, $f_x, f_y$ are bounded on $A$. How does one actually show that $f$ is uniformly continuos on $A$? I imagine that I have to use Mean Value theorem like that: $$|f(x_0, y_0) - f(x,y)| \leq |f(x_0,y_0) - f(x, y_0)| + |f(x, y_0) - f(x, y)| \leq M|x-x_0| + M|y-y_0|$$ where $M$ is some bound on partial derivatives. How do I get this bound from a convex set?
I have one more question. Why do I need set $A$ to be convex? I don't see how to get a counter-example.
Let $M$ be an upper bound of both $f_x$ and $f_y$. Take $(x_1,x_2),(y_1,y_2)\in A$. For each $t\in[0,1]$, let$$\varphi(t)=f\bigl((1-t)(x_1,x_2)+t(y_1,y_2)\bigr).$$Then\begin{align*}\varphi'(t)&=Df_{(1-t)(x_1,x_2)+t(y_1,y_2)}((y_1,y_2)-(x_1,x_2))\\&=f_x((1-t)(x_1,x_2)+t(y_1,y_2)).(y_1-x_1)+\\&\phantom=+f_y((1-t)(x_1,x_2)+t(y_1,y_2)).(y_2-x_2)\end{align*}and therefore$$|\varphi'(t)|\leqslant M\bigl(|y_1-x_1|+|y_2-x_2|\bigr)\leqslant\sqrt2M\|(y_1,y_2)-(x_1,x_2)\|.$$So, by the mean value theorem,$$|f(y_1,y_2)-f(x_1,x_2)|=|\varphi(1)-\varphi(0)|\leqslant\sqrt2M\|(y_1,y_2)-(x_1,x_2)\|.$$Since this occurs for each two points of $A$, then, given $\varepsilon>0$, if you take $\delta=\frac\varepsilon{\sqrt2M}$, you have$$\|(y_1,y_2)-(x_1,x_2)\|<\delta\implies\bigl|f(y_1,y_2)-f(x_1,x_2)\bigr|<\varepsilon.$$
If $A$ is not convex, you may take $A=\{(x,y)\in\Bbb R^2\mid y\neq0\}$ and $f(x,y)=\operatorname{sgn}(y)$. Then $f_x$ and $f_y$ are bounded, but $f$ is not uniformly continuous.