Non-uniformly continous function I have found a topic here (https://www.sciencedirect.com/topics/mathematics/uniformly-continuous-function) in Theorem 3.3.10 that: There is a continuous function $f$ on $[0, 1]$ which is unbounded, and
therefore not uniformly continuous.
Anyone, please help me to understand it.
The proof you link defines

Let $(I_n)_n$ be a covering of $[0, 1]$ by open intervals $(r_n, s_n)$ (...)

and then it says

But by the property of this cover, it follows that no finite part of it covers [0, 1]

Which is false: every open cover of $[0,1]$ has finite subcover, because $[0, 1]$ is compact. The proof is just wrong.
The proof you link defines

Let $(I_n)_n$ be a covering of $[0, 1]$ by open intervals $(r_n, s_n)$ (...)

and then it says

But by the property of this cover, it follows that no finite part of it covers [0, 1]

Which is false: every open cover of $[0,1]$ has finite subcover, because $[0, 1]$ is compact. The proof is just wrong.