Given a bounded sequence ${X_n}$ with infinite range $[a,b]$, is it possible to prove that the set of all sub-sequential limits is also $[a,b]$.
My reasoning is as follows:
Every point in range $[a,b]$ is a limit point. Take any point from $[a,b]$, e.g. $c$. $c$ is a limit point. Choose $n_1$ such that $d(c,X_{n_1})<1$. Choose $n_i$ where $i>1$, such that $d(c,X_{n_i})<\frac{1}{i}$. We show that there is a sub-sequence converging to $c$.
Update: Following Coffeemath's comment, I realise that the domain is a countable infinity while the range is an uncountable infinity. However, reading Rudin's Principles of Mathematical Analysis proof for theorem 3.6(a), an infinite range $E$ is possible but I wonder how can we ensure that $E$ is a countable infinity especially since we can show that there are limit points in $E$ hence are there not uncountable number of punctured neighbourhoods around the limit point that is non-empty?
Rudin's theorem 3.6(a) - If ${p_n}$ is a sequence in a compact metric space $X$, then some sub-sequence of ${p_n}$ converges to a point of $X$.
His proof -
Let $E$ be the range of ${p_n}$. If $E$ is finite then there is a $p∈E$ and a sequence ${n_i}$ with $n_1<n_2<n_3<...$, such that
$p_{n_1} = p_{n_2} = p_{n_3} = ... = p$.
The subsequence $\{p_{n_i}\}$ so obtained converges evidently to $p$.
If $E$ is infinite, Theorem 2.37 (see below) shows that $E$ has a limit point $p∈X$. Choose $n_1$ so that $d(p,p_{n_1})<1$. Having chosen $n_1,...,n_{i-1}$, we see from theorem 2.20 (below) that there is an integer $n_i>n_{i-1}$ such that $d(p,p_{n_i})<\frac{1}{i}$. Then $\{p_{n_i}\}$ converges to $p$.
Theorem 2.37 - If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
Theorem 2.20 - If $p$ is a limit point of a set $E$, then every neighborhood of $p$ contains infinitely many points of $E$.
