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Definition 1
A topological space $(S, {\cal{T}})$ is compact if any of its open covers contains a finite subcover.
Definition 2
In a topological space $(S, {\cal{T}})$, a set $ A\subset S$ is compact if every open cover of $A$ contains a finite subcover.
Question:
Is it correct that if a topological space $(S, {\cal{T}})$ or a set $A\subset S$ is compact then every sequence in it contains a convergent subsequence?
We can show that every compact space is countably compact (meaning that every countable open cover has a finite subcover). We then show that every countably compact space that is additionally first-countable (meaning that each $x \in X$ has a countable neighborhood basis) is sequential compact (meaning that every sequence $(a_n)_{n \in \mathbb{Z}^+}$ has a convergent subsequence $(a_{n_m})_{m \in \mathbb{Z}^+}$).
The first part is easy. Letting $X$ be a compact space, since all open covers of $X$ contain a finite subcover, it is obviously true that all countable open covers of $X$ contain a finite subcover.
Now suppose $X$ is first-countable. Suppose further that $X$ is not sequentially compact: that is, there exists a sequence of points $( a_n )$ in $X$ that does not have a convergent subsequence. We use this sequence to show that $X$ is not countably compact. First, let $A = \{ a_n \}$ be the collection of points in the sequence $( a_n )$. If $A$ is finite, then it is trivial that $A$ has a convergent subsequence, so suppose that $A$ is infinite. For each $n \in \mathbb{Z}^+$, consider the neighborhood $U_n = X - {\rm Cl} \{ a_k \}_{k \ge n}$ of $a_n$. We claim that the countable collection $\mathcal{U} = \{ U_n \}$ is an open cover of $X$, but that it has no finite subcover.
($\mathcal{U}$ covers $X$) Suppose that $\mathcal{U}$ does not cover $X$; let $x_0 \in X - \bigcup \mathcal{U}$ be an arbitrary element that is not covered. By definition of $\mathcal{U}$ and applying DeMorgan's Laws, this means that $x_0 \in \bigcap_{n \in \mathbb{Z}^+} {\rm Cl} \{ a_k \}_{k \ge n}$. We claim that there is a subsequence $( a_{n_m} )$ of $( a_n )$ that converges to $x_0$. Since $X$ is first-countable, there exists a nested neighborhood basis $B_1 \supseteq B_2 \supseteq \dots$ about $x_0$. For $m = 1$, take an arbitrary element $a_{n_{m = 1}} \in A \cap B_{m = 1}$. Such an element exists because $x_0 \in X - \bigcup \mathcal{U} \subseteq X - U_1 = {\rm Cl} \{ a_k \}_{k \ge 1} = {\rm Cl} A$ and $B_1$ is a neighborhood of $x_0$. For $m > 1$, consider the set $A_m = \{ a_n \}_{n > n_{m - 1}}$, which is non-empty by hypothesis that $A$ is infinite. Take an arbitrary element $a_{n_m} \in A_m \cap B_m$. Such an element exists because $x_0 \in X - \bigcup \mathcal{U} \subseteq X - \bigcup_{p = 1}^{n_{m - 1} + 1} U_p = \bigcap_{p = 1}^{n_{m - 1} + 1} {\rm Cl} \{ a_k \}_{k \ge p} \subseteq {\rm Cl} \{ a_k \}_{k > n_{m - 1}} = {\rm Cl} A_m$ and $B_m$ is a neighborhood of $x_0$. The subsequence $( a_{n_m} )$ converges to $x_0$, since any neighborhood $V$ of $x_0$ contains some basis element $B_{m_V}$ of $x_0$, and $a_{n_m} \in B_{m_V} \subseteq V$ for each $n_m > n_{m_V}$. This contradicts the hypothesis that $( a_n )$ does not have a convergent subsequence, so it must be that such an $x_0 \in X - \bigcup \mathcal{U}$ does not exist.
(No finite subset of $\mathcal{U}$ covers $X$) Suppose there is a finite subset $\{ U_{n_1}, \dots, U_{n_N} \} \subseteq \mathcal{U}$ that covers $X$. Take $n_{N'} > \max \{ n_1, \dots, n_N \}$; clearly $\bigcup_{p = 1}^N U_{n_p}$ does not contain $a_{n_{N'}} \in X$, a contradiction.
The above shows that if a first-countable space $X$ is not sequentially compact, it is not countably compact. Equivalently, if a first-countable space $X$ is countably compact, it is sequentially compact, as desired.
The answer to your follow-up question on metric spaces is in the positive, since all metric spaces are first-countable: for any $x \in X$, the collection $\left\{ B_d \left( x, \frac{1}{n} \right) \right\}_{n \in \mathbb{Z}^+}$ is a countable neighborhood basis about $x$.
