And it is the proof of the above statement.
The proof says there exists $\delta$ such that if $|x-c|<\delta$, then $|f(x)-f(c)|<\varepsilon$ . Then, it says $|f(c)-f(c)|=0<\varepsilon$. My question is if $|x-c|<\delta$, can we say that $x=c$? If not, how can we replace $|f(x)-f(c)|$ by $|f(c)-f(c)|$?
Since you suppose that $\lim_{x\rightarrow c} f(x)=f(c) $, you can replace $f(x)$ by $f(c)$
Also when you think about limit you don't consider limit at a point because it's all about approaching
$|x-c|<\delta \iff -\delta +c\lt x \lt \delta+c$ That means our $x$'s are so close to $c$ so that $f(x)$ could be considered to equal to $f(c)$ when $x$ is in that interval given