Let $X$ be a topological space and let $A \subseteq X$.
Definition 1. $A$ is relatively compact in $X$ if the closure of $A$ in $X$ is a compact subspace.
The above is standard, but there are some prima facie weaker conditions that are sometimes useful.
For example:
Definition 2: $A$ is relatively compact in $X$ if there exists a compact $K$ such that $A \subseteq K$ and $K$ is contained in the closure of $A$ in $X$.
Definition 2'. $A$ is relatively compact in $X$ if there exists a compact $K \subseteq X$ with $A \subseteq K$.
We could even not bother having an actual compact subspace:
Definition 3. $A$ is relatively compact in $X$ if every open cover of $X$ has a finite subset that covers $A$, i.e. given open $U_i \subseteq X$ ($i \in I$) such that $X = \bigcup_{i \in I} U_i$, there is a finite $I' \subseteq I$ such that $A \subseteq \bigcup_{i \in I'} U_i$.
Clearly, definition 1 implies definition 2, and definition 2 implies definition 3.
Furthermore, definition 2 and 2' are equivalent because closed subspaces of compact spaces are compact.
Compact subspaces of Hausdorff spaces are closed, so if $X$ is Hausdorff, then definitions 1 and 2 are equivalent.
Question.
Does definition 3 imply definition 1 in general?
What if $X$ is Hausdorff, or $A$ is open?
Definition 3 is closely related to exponentiability: $X$ is exponentiable if and only if for every open $V \subseteq X$ and every $x \in V$ there is an open $U \subseteq V$ such that $x \in U$ and $U$ is relatively compact in $V$ in the sense of definition 3.
When $X$ is Hausdorff, $X$ is exponentiable if and only if $X$ is locally compact.
Best Answer
Without regularity, definition $3$ does not imply definition $1$ in Hausdorff spaces.
Take a compact $T_2$ space $X=(X,\tau)$ which contains an open subset $G\subseteq X$ for which $X\setminus G$ is infinite and has empty interior.
The family $$\mathcal{B}=\{\{x\}\cup (U\cap G)\mid x\in U\in\tau\}\cup\tau$$ is a base for a topology $\sigma$ on $X$. This topology is evidently not regular, but being larger than $\tau$, it is Hausdorff. Since $\sigma$ is Hausdorff but not regular, it cannot be compact. Note that $G$ inherits the same subspace topology from both $(X,\tau)$ and $(X,\sigma)$.
Now, any subset of $G$ has identical closures in both $(X,\tau)$ and $(X,\sigma)$. In particular, since $G$ is dense in $(X,\tau)$, it is also dense in $(X,\sigma)$. Since $(X,\sigma)$ is not compact, $G$ is not relatively compact in $(X,\sigma)$ as per definition $1$.
On the other hand, suppose that $\mathcal{U}$ is a $\sigma$-open cover of $X$. For each $U\in\mathcal{U}$ let $V_U\subseteq X$ be a $\tau$-open set such that $U\subseteq V_U$ and $V_U\cap G=U\cap G$. Then $\mathcal{V}=\{V_U\mid U\in\mathcal{U}\}$ is a $\tau$-open cover of $X$. Since $(X,\tau)$ is compact, $\mathcal{V}$ contains a finite subfamily $\mathcal{V}'$ which covers $X$.
Put $\mathcal{U}'=\{U\in\mathcal{U}\mid V_U\in\mathcal{V}'\}$. This family might not cover $X$, but if $x\in G$ is contained in $V_U\in\mathcal{V}'$, then $x\in V_U\cap G=U\cap G$, and hence $x\in U\in\mathcal{U}'$. It follows that $\mathcal{U}'$ is a finite subfamily of $\mathcal{U}$ which covers $G$.
We conclude that $G$ is relatively compact in $(X,\sigma)$ as per definition $3$.
To see that this is nontrivial we can take $G=\mathbb{N}$ in $X=\beta\mathbb{N}$, or we can take $X=[0,1]$ and let $G$ be the complement of an infinite convergent sequence.
On the other hand, in the presence of regularity, we can establish that definition $3$ implies definition $1$ for Hausdorff spaces.
First observe that for $A\subseteq X$, under definition $3$, $A$ is relatively compact in $X$ if and only if it is relatively compact in $\overline A$. Thus the claim reduces to the statement that for a regular Hausdorff space $X$, a dense subset $A\subseteq X$ is relatively compact in the sense of definition $3$ if and only if $X$ is compact.
Now, recall ([1, Exercise 3.12.5, pg.222]) that a Hausdorff space is H-closed if it is a closed subspace of any other Hausdorff space in which it embeds. We have the following
Putting these facts together yields the following.
[1]: Ryzard Engelking, General Topology, Revised Edition, Helderman Verlag Berlin, (1989).