Given $f:\Bbb R\to \Bbb R$ and $a\in \Bbb R$ define $F(x)=\dfrac{f(x)-f(a)}{x-a}$ for $x\neq a$.Prove that if $f$ is differentiable at $a$ then $F$ is uniformly continuous in some punctured neighbourhood of $a$.
Let $f$ be differentiable at $a$ and let the derivative of $f$ at $a$ be $f^{'}(a)$.
Then given $\epsilon>0\exists \delta_0 >0$ such that $|\dfrac{f(x)-f(a)}{x-a}-f^{'}(a)|<\epsilon $ whnever $0<|x-a|<\delta_0$.
I take the punctured neighbourhood to be $B(a,\delta_0)$.
Now if we choose $x,y\in B(a,\delta_0)$ then $|\dfrac{f(x)-f(a)}{x-a}-f^{'}(a)|<\epsilon $ and $|\dfrac{f(y)-f(a)}{y-a}-f^{'}(a)|<\epsilon $
Now to show that $F$ is uniformly continuous in $B(a,\delta_0)$,take $\epsilon>0$ and we choose $\delta=\delta_0/2$
then $|F(x)-F(y)|=|(\dfrac{f(x)-f(a)}{x-a}-f^{'}(a))-(|\dfrac{f(y)-f(a)}{y-a}-f^{'}(a)|)|<|\dfrac{f(x)-f(a)}{x-a}-f^{'}(a)|+|\dfrac{f(y)-f(a)}{y-a}-f^{'}(a)|<2\epsilon$
whenever $0<|x-a|<\delta_0$.
Hence $F$ is uniformly continuous.
But my friend said he has found a counter-example where he has taken $f(x)=x^21_{\Bbb Q}$.
Where is the fault in my argument?Please help me to find out why my proof is not correct?
The only flaw in your argument is at the very end: You need the inequality whenever $0<|x-y|<\delta$, and your argument does not give you that. But actually, you started going wrong a bit earlier, when you said to choose $x,y\in B(a,\delta_0)$. But that will constrain you to a neighbourhood of $a$, smaller and smaller as $\epsilon$ shrinks, and that spells disaster for your proof idea.
Edited to add: You have proved, quite correctly, that given $\epsilon>0$, there is a $\delta>0$ so that $|F(x)-F(y)|<2\epsilon$ whenever $|x-a|<\delta$ and $|y-a|<\delta$ (you left out the last part).
But that is not the definition of uniform continuity, so you haven't shown that. For uniform continuity, you need to show that given $\epsilon>0$, there is a $\delta>0$ so that $|F(x)-F(y)|<2\epsilon$ whenever $|x-y|<\delta$. But since $|x-y|<\delta$ does not imply either one of $|x-a|<\delta$ or $|y-a|<\delta$, much less both at the same time, your proof does not work.
Edit the second, in response to the edit of the question on June 9:
Note that you are supposed to show that $F$ is uniformly continuous in somme punctured neighbourhood of $a$. That means: There exists such a punctured neigbourhood, say $U$, so that, for every $\epsilon>0$, there is some $\delta>0$ so that $x,y\in U$ and $|x-y|<\delta$ imply $|f(x)-f(y)|<\epsilon$.
Note the order: The neighbourhood is chosen first. You have chosen a neighbourhood depending on $\delta$, and hence on $\epsilon$. But that is an error of logic. Choosing a smaller $\epsilon$ may, in principle, force you to pick a smaller $\delta$, thus changing your neighbourhood.
For reference, here is a correct proof. As you can see, a somewhat bigger gun is needed.
First, note that the differentiability at $a$ implies $\lim_{x\to a}F(x)=f'(a)$. This means that if we extend the definition of $F$ by setting $F(a)=f'(a)$, then $F$ is continuous at $a$. By the (omitted) assumption that $f$ is continuous, it now follows that $F$ is continuous everywhere.
Next, recall that a continuous function on a closed and bounded interval is uniformly continuous. So $F$ is uniformly continuous on $[a-1,a+1]$, say. But then it is also uniformly continuous on any subset of that interval, in particular, on the punctured neighbourhood $(a-1,a+1)\setminus\{a\}$.