Second countability is enough. The following proof is adapted from the proof of Proposition 1.6.23 from the book Convergence Structures and Applications to Functional Analysis by R.Battie and H.-P Butzmann, a fantastic book on convergence spaces.
Proof: Compact $\Longrightarrow$ Sequentially Compact: Assume that $X$ is compact. Take a sequence $\xi:\mathbb N\to X$ in
$X$ and an ultrafilter $\mathcal{G} \supseteq<\xi>.$ Here $<\xi>$ is the associated filter of the sequence $ \xi.$ Then $\mathcal{G}$ converges to a point $x \in X$ and thus $N_x\subset\mathcal G,$ where $N_x$ is the neighborhood filter at $x.$ Since $X$ is second countable, $X$ is also first countable, thus $N_x$ has a countable basis, say $\mathcal F:=\{F_1,\cdots,F_n,\cdots\}.$ Then $F_{n} \cap\left\{\xi_{i} : i \geq k\right\}$ is not empty for all $n, k$ and so one can choose a subsequence
$\xi^{\prime}$ of $\xi$ such that $<\xi^{\prime}>\supseteq \mathcal{F} .$ This means that $\xi^{\prime}$ converges to $x .$ Therefore $X$ is sequentially compact.
Sequentially Compact $\Longrightarrow$ Compact:
Let $\mathcal B$ and $\mathcal C$ be a countable basis and open cover for $X,$ respectively. Define a collection by $$\mathcal B_0:=\{U\in \mathcal B|\ U\ \text{is contained in a member of $\mathcal C$}\},$$ then
we claim that $\mathcal B_0\cap F\neq\emptyset$ for each convergent filter $F.$ In fact, since $\mathcal C$ is an open cover, for each $x\in X$ there is some $V\in\mathcal C$ with $x\in V.$ Since $\mathcal B$ is a basis for $X,$ there is some $U\in\mathcal B$ with $x\in U\subset V,$ thus $U\in\mathcal B_0\cap N_x.$ Hence $\mathcal B_0\cap N_x\neq\emptyset,$ therefore $\mathcal B_0\cap F\neq\emptyset$ for each filter $F$ convergent to $x,$ which proves our claim.
Let $\mathcal B_0=\{B_1,B_2,B_3,\cdots\},$ then we claim that there must be a $n$ such that $\bigcup_{i=1}^n B_i=X.$ We use proof by contradiction. Assume that $\bigcup_{i=1}^n B_i\subsetneqq X$ for all $n\in\mathbb N,$ then for each $n$ choose $\xi_n\in X-\bigcup_{i=1}^n B_i,$ then $\xi_n$ is a sequence in $X,$ thus has a convergent subsequence $\eta_k=\xi_{n_k}.$
Since $\langle \eta_k\rangle$ is a convergent filter, it follows that $\mathcal B_0\cap\langle \eta_k\rangle\neq\emptyset,$ thus $\eta_k$ eventually falls into some $B_n.$ Thus there is some $N\in\mathbb N$ such that $\eta_k\in B_n$ for all $k\geq N.$ However, let $N':=\max(N,n),$ then we have $\eta_k=\xi_{n_k}\in X-\bigcup_{i=1}^{n_k}B_i\subset X-\bigcup_{i=1}^{k}B_i$ for all $k\geq N',$ which is a contradiction. Thus there must be some $n_0$ with $\bigcup_{i=1}^{n_0}B_i=X.$
Note that for each $i$ there is a $C_i\in\mathcal C$ with $B_i\subset C_i,$ it follows that $C_1, C_2,\cdots,C_{n_0}$ is a finite subcover for $\mathcal C,$ thsu $X$ is compact, which completes the proof.
Best Answer
This is equivalent to local compactness.
If a topological space satisfies your property, then a singleton point set $\{x\}$ is compact, and hence there must be a relatively compact set $T$ whose interior contains $x$ (in fact, it contains $\overline{\{x\}}$, which may be strictly larger). The closure of $T$ is a compact neighbourhood of $x$.
If, on the other hand, if we have locally compact space and a relatively compact set $S$, then $\overline{S}$ is compact. For each point $s \in \overline{S}$, there is an open neighbourhood $\mathcal{N}_s$ of $s$ that is compact. Note that the interiors of these neighbourhoods cover $\overline{S}$, hence there must be a finite-subcover of $\overline{S}$. Unioning this finite subcover yields an open, relatively compact neighbourhood containing $\overline{S}$, establishing equivalence.