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: 
To prove (1), please look up [Bolzano-Weierstrass theorem](https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem)

For (2), note that any subsequence of your original bounded sequence is going to be bounded as well. Therefore, we know such subsequence will, itself, contain a converging subsequence (by (1)). Since there are an infinite number of distinct subsequences of the original sequence (if the sequence is not constant after a certain index), there are an infinite number of converging subsequences.
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 $a\_n$.