Prove that a closed subset of a compact set is compact using only the sequential definitions. I want to prove the following statement.
A closed subset of a compact set is compact.
But I want to use only the sequential definitions.
My attempt, take $B$ a compact set and $A \subset B$ closed subset, then every sequence $x_n \in B$ has a convergent subsequence $x_{n_k} \to x$ where $x \in B$.
Take a sequence $y_n \in A$, since $A$ is a subset of $B$ then $y_n$ has a convergent subsequence $y_{n_k} \to y$ where $y \in B$.
I want to prove that $y \in A$, since $A$ is closed then it contains all of its limit points therefore $y \in A$, proving that $A$ is compact.
Is my proof correct?
 A: In general, it's not enough that "every sequence has a converging subsequence" to say "$X$ is compact" (a space where every sequence has a converging subsequence is sequentially compact), what you need is the notion of net. A net in a space $X$ is merely a function $f:D\rightarrow X$, where $D$ is a partially ordered, directed set (that is, $D$ has a partial order and for every $d,d'\in D$ there is $e\in D$ such that $e\geqslant d,d'$). A special case is a sequence, where $D=\mathbb N$ with its usual order. A space $X$ is compact if and only if every net (not just a sequence) has a converging subnet. A subnet of a net $f:D\rightarrow X$ is a net in the form $f\circ j:D'\rightarrow X$, where $j$ is a monotone map $D'\rightarrow D$ between directed sets and for every $d\in D$ there is $d'\in D'$ such that $d\leqslant j(d')$. A net $f:D\rightarrow X$ converges to $x\in X$ if for every neighborhood $U$ of $x$ there is $d\in D$ such that $$f(d')\in U\forall d\leqslant d'$$
If $X$ has some other property (for example, when $X$ is a metric space), then sequences are enough to check that $X$ is compact, but not in general. It is the same for continuous functions: A function $\phi:X\rightarrow Y$ is continuous if and only if for every net $f:D\rightarrow X$ (with $D$ any directed set), if $x$ is a limit point of $f$, then $f(x)$ is a limit point of $\phi\circ f$. Again, there are cases where $f$ is continuous if and only if the previous condition is satisfied just for sequences.
Nonetheless, your proof is correct and the difference with using nets instead of sequences is basically null (the overall scheme remains unchanged: you consider a net $f:D\rightarrow Y$, which is a net in $X$, so it has a converging subnet $f':D'\rightarrow X$ whose limits belong to $Y$ since $Y$ is closed in $X$).
A: Such a proof is fine in a context where "$X$ is compact" is equivalent to "every sequence in $X$ has a convergent subsequence with limit in $X$".
This is the case in metric and metrisable spaces.
In that context the proof is fine, and if we're in a metric space we gan go back too because limits are unique:
let $Y$ be a closed subset of a compact $X$. If $(y_n)_n$ is a sequence in $Y$ it has a convergent subsequence $(y_{n_k})_k$ with limit $x \in X$, but then closedness of $Y$ plus $y_{n_k} \in Y$ for all $k$ implies $x \in Y$ and that same subsequence then shows the compactness of $Y$.
and if $Y$ is a compact subset of $X$, let $p \in \overline{Y}$ and we have a sequence $(y_n)_n$ in $Y$ that converges to $p$. As $Y$ is compact, there is some $y_0 \in Y$ such that $(y_{n_k})_k \to y_0$ for some subsequence. But clearly $(y_{n_k})_k \to p$ as well and by unicity of sequence limits $y_0 = p$ and so $p \in Y$ and hence $\overline{Y} \subseteq Y$ and so $Y$ is closed.
In a general sequential Hausdorff space $X$ we thus have by this last proof:

A sequentially compact subspace $Y$ of $X$ is closed in $X$.

