As stated in the title, given a poset $(S,\leq)$ I think it's trivial that an unbounded subset $A \subseteq S$ is cofinal, but does the opposite implication hold?
Definition 1. A subset $X$ of a poset $(S,\leq)$ is said to be cofinal if: $\forall s \in S \; \exists x \in X \; s \leq x$.
Definition 2. A subset $X$ of a poset $(S,\leq)$ is said to be bounded: $\exists s \in S \; \forall x \in X \; x \leq s$.
Neither implication holds.
In $(\mathbb{N}_+, \mid)$ the subset $\{ 2^n : n \in \mathbb{N} \}$ is unbounded, but not cofinal.
In $(\mathcal{P}(\mathbb{N}), \subseteq)$ the subset $\{ \mathbb{N} \}$ is cofinal, but bounded.