For a function ($f:[a,b] \to \Bbb R$) to satisfy Rolle's theorem it must be continuous in the interval $[a,b]$, differentiable in $(a,b)$ and $f(a)=f(b)$. I don't really understand what's the need of function being continuous at $x=a$ and at $x=b$.
Because even if the functions are not continuous at $ a,b$ then also I can get a '$c$' in $(a,b)$ such that $f'(c)=0$ by just replacing the third condition written above with $$\lim_{x\to a}f(x)=\lim_{x \to b}f(x)$$
So is there any specific reason to define continuity at the end points?
In comments, you mentioned you wanted to know why continuity at the endpoints is needed for the proof.
I will show why that is with two counterexamples. It's all about the extreme value theorem.
By the way, it's worth also noting that $f(a)=f(b)$ does not mean $\lim_{x\to a}f(a)=\lim_{x\to b}f(x)$. The definition of limit actually means that neither limit actually involves the values $f(a)$ or $f(b)$ so there is no reason to expect it to be true. And it is false: take $$f:[0,1]\to\Bbb R,\,\,x\mapsto \begin{cases}0&x=0,1\\1-2x&0<x<1\end{cases}$$
The limit as $x\to a=0$ is $1$ but the limit as $x\to b=1$ is $-1$.
I will use this same $f$ for my first counterexample. So, $f$ is differentiable on $(0,1)$ and $f(0)=f(1)$. On the interior, $f'(x)=-2$ always - it is never zero. So what went wrong? The standard proof of Rolle's theorem begins with:
Notice that $\sup_{x\in[0,1]}f(x)=1$ and $\inf_{x\in[0,1]}f(x)=-1$ but $f$ is never equal to $1$ or $-1$. So here the upper bounds do exist - $f$ has a supremum - but it never attains a maximum. So we can't pick $\xi\in(0,1)$ a extremal point and use the usual argumentation to find $f'(\xi)=0$. The openness of $(0,1)$ allows for $f$ to be always decreasing without ever reaching its lower limits. The closedness of $[0,1]$ eliminates/helps to eliminate that possibility, which is why it is important that $f$ be continuous on all of $[0,1]$ in Rolle's theorem.
Now for another example. Let: $$f:[0,1]\to\Bbb R,\,x\mapsto\begin{cases}0&x=0,1\\\cot\pi x&0<x<1\end{cases}$$
$f$ is again differentiable on the interior and $f(0)=f(1)$. But $f$ is not continuous at the endpoints, and in fact $f$ does not have any upper or lower bounds! The extreme value theorem fails even more significantly in this example. $f'$ is never zero. A lack of continuity at the endpoints leaves open the possibility that $f$ is unbounded, which is a different problem to the nuance between being bounded but not attaining that bound.
The extreme value theorem is so useful because it gives both a bound and an attaining of that bound. Weaken the hypotheses at your peril!