Question: [Extensions of measurable functions](https://math.stackexchange.com/q/76568)
Let $(X,F)$ be a measurable space. $U \subset X$, which is not necessarily measurable. $f$ is a real-valued measurable function defined on $(U,F\cap U)$, then we can extend $f$ to a measurable function on $X$.
The answer in the link above makes me confused, I don’t know because $U$ must be measurable.

Answer: 
As many pointed (1) is the classical Weierstrass---Bolzano theorem.

For (2) your doubts are coming from confusing a sequence with the the values it takes.

Consider the sequence $a\_{n}=2$. Then, $a\_{2n}$ and $a\_{3n}$ are (typically considered) different subsequences of $a\_n$, even if they are the same sequence.

They are different subsequences because they have different indeces. Formally a subsequence is $a\_{\phi(n)}$ where $\phi : \mathbb N \to \mathbb N$ is a strictly increasing function, and different functions give different subsequences.
The issue is that we are often a bit careless when it comes to this, thinking of a subsequence as the values it takes.

For (2) note that if $a\_n$ has a convergent subsequence $a\_{\phi(n)}$ then any subsequence of $a\_{\phi(n)}$ is a cnvergent subsequence of $a\_n$.
Mere bhai , you can only do this , select a element and repeat it infinite times only when range set is finite.

The idea of this proof is that , if a sequence is bounded then it's means it's range set is bounded . And you may also know about that every bounded subset of $\mathbb{R}^k$ lies in compact subset of $\mathbb {R}^k$ and its a fact that sequence in compact matric space must have a sub sequence in it.