** There is a subs. $\displaystyle \{x_{n_k}^{(1)}\}$ converging to $\displaystyle L_1\Longrightarrow\,\,\,for\,\,\,\epsilon=1\,\,\ex ists\,K_1\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{n_k}^{(1 )}-L_1|<1\,\,\,\forall\,\,n_k^{(1)}>K_1$ . Choose now $\displaystyle n_1:=n^{(1)}_{K_1+1}$ .
** There is a subs. $\displaystyle \{x_{n_k}^{(2)}\}$ converging to $\displaystyle L_2\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 2\,\,\exists\,K_2\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(2)}-L_2|<1\slash 2\,\,\,\forall\,\,n_k^{(2)}>K_2$ . Choose now $\displaystyle n_2:=n^{(2)}_{K_2+h_2}$ in such a way that $\displaystyle n_2>n_1$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_2 \,\,s.t.\,\,\,K_2+h_2>K_1+1$ !)
** There is a subs. $\displaystyle \{x_{n_k}^{(3)}\}$ converging to $\displaystyle L_3\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash 3\,\,\exists\,K_3\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(3)}-L_3|<1\slash 3\,\,\,\forall\,\,n_k^{(3)}>K_3$ . Choose now $\displaystyle n_3:=n^{(3)}_{K_3+h_3}$ in such a way that $\displaystyle n_3>n_2$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_3 \,\,s.t.\,\,\,K_3+h_3>K_2+h_2$ !)
Continue inductively in the above way, and in the m-th step:
** There is a subs. $\displaystyle \{x_{n_k}^{(m)}\}$ converging to $\displaystyle L_m\Longrightarrow\,\,\,for\,\,\,\epsilon=1\slash m\,\,\exists\,K_m\in\mathbb{N}\,\,\,s.t.\,\,\,|x_{ n_k}^{(m)}-L_m|<1\slash m\,\,\,\forall\,\,n_k^{(m)}>K_m$ . Choose now $\displaystyle n_m:=n^{(m)}_{K_m+h_m}$ in such a way that $\displaystyle n_m>n_{m-1}$ (we can do this because we've an infinite subsequence here and thus there exists $\displaystyle h_m \,\,s.t.\,\,\,K_m+h_m>K_{m-1}+h_{m-1}$ !)
Take it from here...
Tonio