Assume that there exists for the sequence $t_1, t_2, \dots$ a sequence of positive numbers $\alpha_1,\alpha_2,\dots$, converging to $0$, for which $$t_{n+1}>t_n-\alpha_n \ \text{for all} \ n\in \mathbb{N}.$$ Then the numbers $t_1,t_2,\dots, $ are everywhere dense between their lowest and highest limit points.
Solution: Suppose that $l=\liminf \limits_{n\to \infty} t_n$ and $L=\limsup \limits_{n\to \infty} t_n$.
It easy to check that the case $l=L$ is obvious. It has three subcases: $l=L=\pm \infty$ and $l=L$ is a real number.
So we suppose that $l<L$ and $l,L\in \mathbb{R}$ (the case when at least one of them is infinity can be proven in the same way just with slight modifications).
So our goal to show that $\{t_n: n\in \mathbb{N}\}$ is dense in $[l,L]$. It is enough to show its density on an open interval $(l,L)$.
Suppose that $\xi\in (l,L)$ and $\epsilon>0$ be given. Consider $$\delta:=\frac{1}{2}\min \{\xi-l, L-\xi, \epsilon\}>0.$$
Also $\exists N_0$ such that if $n\geq N_0$ then $0<\alpha_n<\delta/2$.
By definition of $l$ and $L$ we can find $N>M>N_0$ such that $|t_M-L|<\frac{\delta}{2}$ and $|t_N-l|<\frac{\delta}{2}$. Then I claim that the following statement is true.
Statement: At least one of the $\{t_M,\dots,t_N\}$ belongs to the interval $(\xi-\delta, \xi+\delta)$.
Proof: Suppose that is not true. Consider the following set $$\mathcal{C}:=\{i\in \{M\dots,N\}: t_i\geq \xi+\delta\}.$$ We note that $M\in \mathcal{C}$ because $\delta\leq \frac{1}{2}(L-\xi)$.
Then $\exists \max \mathcal{C}=p$. Then $p+1\notin \mathcal{C}$ and hence $t_{p+1}<\xi+\delta$. Since we assumed that none of the elements of $\{t_M,\dots, t_N\}$ belongs to $(\xi-\delta,\xi+\delta)$ then $t_{p+1}<\xi-\delta$. Hence $t_p-t_{p+1}\geq 2\delta$.
Since $p\geq M>N_0$ hence $t_{p+1}-t_p>-\alpha_p>-\delta/2$. Combining these two inequalities we obtain that $\delta/2>t_p-t_{p+1}\geq 2\delta$ which is contradiction. Since the statement is valid then $\exists k$ such that $|t_k-\xi|<\delta<\epsilon.$
As I told above this proof can be easily adapted if at least one of the $l$ or $L$ are infinite.
The proof seems correct to me but I was wondering is everything alright here? Thanks a lot for attention!
EDIT: Usually when we say that $A$ is dense in $X$ we mean $A$ to be subset of $X$.
But I notice that in this case $\{t_n:n \in \mathbb{N}\}$ may not be subset of $[l,L]$ because some elements of $t_n$ may be outside of $[l,L]$. Can anyone explain it to me, please?