Let $\{ t_n \}$ be a bounded sequence on the real line. If all the convergent subsequence $\{ t_{n_k} \}$ of $\{ t_n \}$ converge to same value $t_0$, then can I say that $\lim_{n \to \infty} t_n = t_0$ ?

I would argue as follows... Since $\{t_n\}$ is bounded, there exists $R>0$ such that $t_n\in [-R,R]$ for every $n\in\mathbb N$. Then pick a subsequence $\{t_{n_j}\}_j$. As $[-R,R]$ is sequentially compact, there exists a sub-subsequence $\{t_{n_{j_k}}\}_k$ which converges. As this is a particular convergent subsequence of ${t_n}$, then it must converges to some $\bar t$, which is unique for all these subsequences by hypothesis. As $\mathbb R$ is Hausdorff, it follows that $\{t_n\}$ converges to $\bar t$.
We have used the well known fact that if we have a sequence, and we know that from every subsequence we can extract a convergent sub-subsequence, and all of these sub-subsequences have the same limit $t_0$, then also the sequence converges to $t_0$.
Since the sequence $(t_n)_{n \in \mathbb{N}}$ is bounded,has a convergence subsequence to $t_0$.
If $t_n \not \to t_0$, then $\exists \epsilon >0$ s.t. $\forall k \in \mathbb{N}$ exist $n_k>k$ with $|t_{n_k}-t_0|>\epsilon.$ You can take $n_k$ to be strictly increasing. So the subsequence $(t_{n_k})_{k \in \mathbb{N}}$ is bounded and does not have subsequence converging to $t_0$. This is a contradiction.