While Proving Rolle's Theorem we have if $M$ and $m$ are Supremum and Infimum values of $f(x)$ in $\left[a ,\:\: b\right]$ and $M \ne m$
Then Supposing $f(a) \ne M$, Then $f(b) \ne M$ and Now there exists $c \in (a, b)$ such that $$f(c)=M$$
Now in the Neighborhood of $c$ there exists $\epsilon \gt 0$ such that
$$f(c+\epsilon)< f(c)$$ $\implies$
$$f'(c)=\lim_{\epsilon \to 0}\frac{f(c+\epsilon)-f(c)}{\epsilon} \lt 0 \tag{1}$$ Similarly
in the Neighborhood of $c$ there exists $\epsilon \gt 0$ such that
$$f(c-\epsilon)< f(c)$$ $\implies$
$$f'(c)=\lim_{\epsilon \to 0}\frac{f(c)-f(c-\epsilon)}{\epsilon} \gt 0 \tag{2}$$
Now since $(1)$ and $(2)$ are contradicting we have
$$f'(c)=0$$
Now my Doubt is in my book, there is also equal to sign in both $(1)$ and $(2)$
How can there be equal to sign?
I don't quite understand your question, but I believe the confusion lies in the following. The correct version of $(1)$ is $$f'(c)=\lim_{\epsilon \to 0}\frac{f(c+\epsilon)-f(c)}{\epsilon} \leq 0,$$ and the correct version of $(2)$ is $$f'(c)=\lim_{\epsilon \to 0}\frac{f(c+\epsilon)-f(c)}{\epsilon} \geq 0.$$
For a computational example to witness this firsthand, you might check out $f(x) = -(x-1)^2 + 1$ on $[0, 2]$. Sometimes working out an example clears up any remaining confusions.