ขอใช้ภาษาอังกฤษนะครับ

Suppose WLOG that $f$ is increasing.

Fix $a\in\mathbb{R}$. Let $\epsilon>0$ be given.

Then there are $b,c$ such that

$f(b)=f(a)+\frac{\epsilon}{2}$ and $f(c)=f(a)-\frac{\epsilon}{2}$.

Observe that $b>a>c$.

Let $\delta_1=b-a>0$ and $\delta_2=a-c>0$.

Then $f(a+\delta_1)=f(b)=f(a)+\frac{\epsilon}{2}<f(a)+\epsilon$ and

$f(a-\delta_2)=f(c)=f(a)-\frac{\epsilon}{2}>f(a)-\epsilon$.

Choose $\delta=\min{\{\delta_1,\delta_2\}}$.

If $|x-a|<\delta$ then

$a-\delta<x<a+\delta$,

$a-\delta_2\leq a-\delta < x < a+\delta\leq a+\delta_1$.

Thus $f(a-\delta_2)<f(x)<f(a+\delta_1)$,

$f(a)-\epsilon<f(a-\delta_2)<f(x)<f(a+\delta_1)<f(a)+\epsilon$.

Hence $|f(x)-f(a)|<\epsilon$.

Therefore $f$ is continuous at $a$.

This completes the proof. :yum: