If the infinite sequence$\ (b_{n}) \ $converges to $\ b,\ $are$\ t_{n} \ =inf\{b_{n},b_{n+1},b_{n+2},...\} \ $and$\ g_{n} \ =sup\{b_{n},b_{n+1},b_{n+2},...\} \ $subsequences of$\ (b_{n}) \ $?
Recall the following:
If we let$\ (b_{n}) \ $be an infinite sequence of real numbers, and let $\ n_{1}<n_{2}<n_{3}<n_{4}<... \ $be an increasing sequence of natural numbers, then the sequence:
$$\ (b_{n_{1}},b_{n_{2}},b_{n_{3}},b_{n_{4}},...) \ $$
is called a subsequence of$\ (b_{n}) \ $and is denoted by$\ (b_{n_{k}}), \ $where$\ k\in\mathbb{N}^{+} \ $indexes the subsequence.
(Note: the above definition of a subsequence is taken from Stephen Abbott's Understanding Analysis)
If $\ (b_{n}) \ $converges to$\ b \ $, then:
(1)$$\ \lim_{n\to \infty}supb_{n} \ =\lim_{n\to \infty}g_{n}=b $$
(2)$$\ \lim_{n\to \infty}infb_{n} \ =\lim_{n\to \infty}t_{n}=b $$
As we know, all subsequences of a convergent sequence converge to the same limit as the convergent sequence - thus, by (1) and (2) our question is leaning more towards being correct...
The key point here, is that the supremum/infimum of a set need not belong to that set, and the first counterexamples to this emerge from the very simple sequence $\frac 1n$. For example, the sequence $t_n$ for $\frac 1n$ is just $0,0,0....$ which is not a subsequence.
For a similar counterexample with the suprema, take $1−\frac 1n$, and you can sort of alternate these sequences with little modifications, to get an example where both $t_n$ and $g_n$ are not subsequences of $b_n$.
Nextly, if there is a sequence, say $a_n$ which converges to $a$,then what is definitely true, is that every subsequence of $a_n$ converges to $a$ as well. What is not true, is that if there is some other sequence converging to $a$, then it needs to be related to $a_n$ in a subsequential manner. So, the intuition you are attempting to use to get to your conclusion is unfortunately incorrect, but the statements you have written are true, and significant in their own right.