Can a subsequence repeat a term finitely many times? Suppose $\{a_n\}$ is a sequence. Let $f(n)$ be an increasing function(not strictly) such that $f$ takes each of its values finitely many times, then wouldn't $\{a_{f(n)}\}$ have the same limit as $\{a_n\}$ (assuming $\{a_n\}$ converges)? Then shouldn't $\{a_{f(n)}\}$ be considered a subsequence of $\{a_n\}$?
UPDATE: Edited to clarify what I meant.
 A: The standard definition for subequence requires $f(n)$ to be strictly increasing. See this for example. 
If I understand you correctly, you would allow $f$ to be something like
$$f(n) = \begin{cases} 1 & \text{if $1 \le n \le 5$} \\ n & \text{ if } 5 \lt n  \end{cases} $$
For this $f$, the sequence $a_{f(n)}$ wouldn't be considered to be a subsequence of $a_n$ per the usual definition; but it is still a sequence, and it does indeed converge to the same limit as $a_n$. 
And this of course holds in general; by repeating some of the terms, you are just "delaying" the $N$ after which every term will be within $\epsilon$ of the limit.
I think you probably understand the idea, but just to address your very last sentence: the $a_{f(n)}$ is not considered a subsequence because and only because $f$ is not strictly increasing. 
A: No. A subsequence is not about "choosing terms from the sequence", but about choosing a subset of indices. So if your sequence is $$(a_1,a_2,\ldots),$$ then a subsequence is determined for example by choosing the even indices: $$(a_2,a_4,a_6,\dots),$$ or the multiples of 7, $$(a_7,a_{14},a_{21},\ldots),$$ or in general any strictly increasing function $\gamma:\mathbb N\to \mathbb N$, and the subsequence is given by $$(a_{\gamma(1)},a_{\gamma(2)},\ldots).$$
A: Let $S= \{a_i\}$ be the sequence $\{\frac 1n\}$ so $S = \{1,\frac 12,\frac 13,\frac 14, ....\}$.
Let $f(1)= 1,f(2) = 1$ but $f(k) = k$ for $k > 2$.
Then $T = \{a_{f(i)}\} = \{1,1,\frac 13, \frac 14\}$.
Then $T$ is not a subsequence of $S$ because $T$ has two terms equal to $1$ but $S$ only has one.
However, yes, $\lim a_i = \lim f_{f(i)} = 0$.  (Assuming $\lim f(n) = \infty$)
But why should that imply $T$ is subsequence of $S$????
Lot's of sequences converge to the same limit.  Consider $S= \{2 + \frac 1k\}$ and $T = \{2- \frac 1k\}$.  Then $S = \{3,2\frac 12, 2\frac 13,2\frac 14, ....\}$ and $\lim S = 2$.  But $T = \{1, 1\frac 12, 1\frac 23, 1\frac 34,....\} $ and $\lim T = 2 = \lim S$.
But no-one can claim one is the subset of the other. 
