I take it that your edit is the question you wanted to ask, i.e.
Is there a countable subset S of T such that for every element $t\in T$, there exists $s\in S$ such that $s\geq t$?
So let $T$ be uncountable.
If $T$ is bounded above then it has a supremum $\eta$. If $\eta \in T$ then let $S = \{ \eta \}$. Otherwise, let $S = \{ \eta_n\, :\, n \in \mathbb{N} \}$ where $(\eta_n)_{n=1}^{\infty}$ is some strictly increasing sequence in $T$ converging to $\eta$.
If $T$ is unbounded then let $S = \{ \eta_n\, :\, n \in \mathbb{N} \}$, where $(\eta_n)_{n=1}^{\infty}$ is some strictly increasing sequence in $T$ which tends to $\infty$.
In any case, by construction, for each $t \in T$ there exists $s \in S$ such that $s \ge t$, as you should verify; and $S$ is clearly countable.
If you require $S$ to be countably infinite then, in the case $T$ bounded and $\sup T = \eta \in T$ given above, let $S'$ be any countably infinite subset of $T$ (e.g. constructed as in Ayman Hourieh's answer) and then let $S = S' \cup \{ \eta \}$.