How shall we prove that the converse statement of “epsilon-delta definition of continuity” is true

Question

The epsilon-delta definition of continuity:

A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x–x_0| < \delta$ then $|f(x)–f(x_0)| < \epsilon$

Now from this, how shall we deduce that the converse statement is true?

Converse statement:

If a function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$, then for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x–x_0| < \delta$ then $|f(x)–f(x_0)| < \epsilon$

Answer 1

Long comment

See T.Tao, Analysis, I, page 227 :

Definition 9.4.1 (Continuity). Let $X$ be a subset of $\mathbb R$, and let $f : X → \mathbb R$ be a function. Let $x_0$ be an element of X. We say that $f$ is continuous at $x_0$ iff we have :

$\lim_{x → x_0;x∈X} f(x) = f(x_0).$

Thus, there is nothing to prove.