A semi non-computable function? Maybe it's meaningless to ask this question. I'm asking,because I don't want have a question mark in my head.

Let, $f(n)$ be a non-computable function , $f:\mathbb{N}\longrightarrow \left\{0,1\right\}$. In other words,
$$f(n):=\left\{a_n\right\}_{n\in\mathbb{N}}$$

So here is my question:

Is it possible to choose an infinite computable subsequence from the sequence $\left\{a_n\right\}_{n\in\mathbb{N}}$ ?

I'm confused at this point. If this is possible, we get semi non-computable sequence:
For example:

$a_n:=\left\{a_1,a_2,0,a_4,a_5,1,a_7,a_8,0,a_{10},a_{11},1\cdots\right\}$

A subsequence $\left\{a_1,a_2,a_4,a_5,a_7,a_8,a_{10},a_{11},\cdots\right\}$ is obviously non-computable. But the sequence, $\left\{0,1,0,1,0,1,\cdots\right\}$ is computable.
So, we can define a semi non-computable function/ infinite sequence.
Is that what I say is nonsense?
 A: Any infinite sequence of $0$s and $1$s can be broken into at most two computable subsequences, namely the $0$s and the $1$s (the "at most" is to handle the case in which only finitely many $0$s appear, or only finitely many $1$s appear - note that in this case the whole sequence is computable).
However, while a subsequence $\beta$ of a sequence $\alpha$ may be computable, knowing how $\beta$ fits inside $\alpha$ may still be quite complicated, and so the mere existence of such a $\beta$ doesn't actually limit the complexity of $\alpha$ at all.

Incidentally, there are infinite sequences of naturals which have no computable subsequences. The most natural (hehe) of these is probably the Busy Beaver sequence, $\mathfrak{B}$: the $n$th term of $\mathfrak{B}$ is the longest time it can take for a Turing machine of size $<n$ to halt on input $0$. Since it's nondecreasing, for any subsequence $\mathfrak{A}$ we have $\mathfrak{A}(n)\ge\mathfrak{B}(n)$ for every $n$; but at the same time, it's a good exercise to show that any sequence which dominates $\mathfrak{B}$ computes $\mathfrak{B}$. So in fact, not only does $\mathfrak{B}$ have no computable subsequences, but every subsequence of $\mathfrak{B}$ is at least as complicated as $\mathfrak{B}$. 
This last property isn't actually rare, though: for every Turing degree d there is an infinite sequence of naturals $\theta$ which has Turing degree d and such that every subsequence of $\theta$ computes $\theta$. This uses a cute trick - replace the sequence $(n_1,n_2,n_3,...)$ with the sequence $([n_1], [n_1,n_2], [n_1,n_2,n_3], ...)$ where "$[a,...,b]$" is the canonical code for the finite sequence $(a,...,b)$. What is rare is the domination property, that any sufficiently fast-growing function computes $\mathfrak{B}$; it turns out that a function is computable from all sufficiently fast-growing functions iff it is hyperarithmetic. But that's a significantly more difficult topic.
