Sequence such that every subsequence can have a different real limit

Just arrange the set of rational numbers in a sequence $\{a_n\}$. Given any real number $L$ and any positive integer $n$ there are infinitely many rationals in $(L-\frac 1 n, L+\frac 1 n)$. Pick $a_{n_1}$ in this interval with $n=1$. Then consider the case $n=2$. You can surely find $n_2 >n_1$ such that $a_{n_2} \in (L-\frac 1 2, L+\frac 1 2)$. Use induction to construct a subsequence $a_{n_k}$ such that $|a_{n_k}-L| <\frac 1 k$ for all $k$. Then $a_{n_k} \to L$.

Take the sequence that sweep the interval $[-1,1]$ by $1/2$ steps, then the interval $[-2,2]$ by $1/2^2$ steps, then the interval $[-n,n]$ by $1/2^n$ steps... and so on.

You’ll be able to prove that every real is a limit point of that sequence.

Rephrasing:

Consider $a_n$, $n=1,2,3,3,....,$ the sequence of rational numbers. Recall that $\mathbb{Q}$ is countable, hence can be written as a sequence $a_n$, $n\in \mathbb{N}$.

Let $L \in \mathbb{R}$.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, we can construct a subsequence $a_{n_k}$ that converges to $L$.