Prove $y = x$ is continuous For every $\epsilon > 0$ there exists a $\delta > 0$ such that $|x - c| < \delta$ implies $|f(x) - f(c)| < \epsilon$.
Start with $|f(x) - f(c)| < \epsilon$ which gives $|x - c| < \epsilon$. We also know $|x - c| < \delta$ but how can we connect $\epsilon$ and $\delta$?

(For future visitors) You can prove it as follows.
For any $x \in \mathbb R$ we have:
$x_n \to x_0 \implies f\left(x_n\right) = x_n \to x_0 = f\left(x_0\right)$
Meaning, if sequence $x_n$ somehow approaches $x_0$, then corresponding sequence $f\left(x_n\right)$ always approaches $f\left(x_0\right)$, which is the definition of continuity. Taking in account that we chose any $x$ from $\mathbb R$ (in the beginning), we've proven that $\lim_{x \to x_0} f\left(x\right) = f\left(x_0\right)$ for any $x \in \mathbb R$. $\blacksquare$
