[proofplan]
We prove both implications directly from the definition of [sequential compactness](/page/Sequential%20Compactness). If $X$ is finite, then every sequence in $X$ has some value occurring infinitely often, and the corresponding constant subsequence converges. Conversely, if $X$ is infinite, we choose a sequence of pairwise distinct points; no subsequence of this sequence can converge in the [discrete topology](/page/Discrete%20Topology), because convergence to a point forces eventual membership in the open singleton of that point.
[/proofplan]
[step:Extract a constant subsequence from any sequence in a finite space]
Assume first that $X$ is finite. Let
\begin{align*}
x: \mathbb{N} \to X
\end{align*}
be a sequence in $X$, and write $x_n := x(n)$ for each $n \in \mathbb{N}$.
If $X = \varnothing$, then there is no sequence $\mathbb{N} \to X$, so the defining condition for sequential compactness is vacuously satisfied. Assume now that $X$ is nonempty and finite. For each $a \in X$, define
\begin{align*}
I_a := \{n \in \mathbb{N} : x_n = a\}.
\end{align*}
The finite family $(I_a)_{a \in X}$ covers $\mathbb{N}$. If every $I_a$ were finite, then $\mathbb{N} = \bigcup_{a \in X} I_a$ would be finite as a finite union of finite sets, contradicting the infinitude of $\mathbb{N}$. Hence there exists $a_0 \in X$ such that $I_{a_0}$ is infinite.
Choose a strictly increasing map
\begin{align*}
k: \mathbb{N} \to \mathbb{N}
\end{align*}
whose image is contained in $I_{a_0}$. Then the subsequence
\begin{align*}
x \circ k: \mathbb{N} \to X
\end{align*}
satisfies $x_{k_m} = a_0$ for every $m \in \mathbb{N}$.
We verify convergence. Let $U \in \tau_{\mathrm{disc}}$ be any open neighbourhood of $a_0$. Since $a_0 \in U$, every term of the subsequence belongs to $U$. Therefore $x_{k_m} \to a_0$. Thus every sequence in $X$ has a convergent subsequence, so $(X, \tau_{\mathrm{disc}})$ is sequentially compact.
[guided]
Assume $X$ is finite. To prove sequential compactness, we must start with an arbitrary sequence and produce a convergent subsequence. Let
\begin{align*}
x: \mathbb{N} \to X
\end{align*}
be a sequence, and write $x_n := x(n)$.
If $X = \varnothing$, then no map $\mathbb{N} \to X$ exists, because $\mathbb{N}$ is nonempty. Thus the statement “every sequence in $X$ has a convergent subsequence” has no counterexample, so it holds vacuously. Now assume $X$ is nonempty.
For each point $a \in X$, define the set of indices at which the sequence takes the value $a$:
\begin{align*}
I_a := \{n \in \mathbb{N} : x_n = a\}.
\end{align*}
These sets cover $\mathbb{N}$ because every term $x_n$ is an element of $X$, so $n \in I_{x_n}$. Since $X$ is finite, this is a finite cover of $\mathbb{N}$ by the sets $I_a$.
At least one of these index sets must be infinite. Indeed, if each $I_a$ were finite, then the finite union
\begin{align*}
\bigcup_{a \in X} I_a
\end{align*}
would be finite. But this union is all of $\mathbb{N}$, contradicting that $\mathbb{N}$ is infinite. Therefore there exists $a_0 \in X$ such that $I_{a_0}$ is infinite.
Because $I_{a_0}$ is an infinite subset of $\mathbb{N}$, we may list its elements in strictly increasing order. Equivalently, choose a strictly increasing map
\begin{align*}
k: \mathbb{N} \to \mathbb{N}
\end{align*}
with $k_m := k(m) \in I_{a_0}$ for every $m \in \mathbb{N}$. The corresponding subsequence is the map
\begin{align*}
x \circ k: \mathbb{N} \to X.
\end{align*}
For every $m \in \mathbb{N}$, the condition $k_m \in I_{a_0}$ means exactly that $x_{k_m} = a_0$. Thus this subsequence is constant with value $a_0$.
It remains to check convergence from the topological definition. Let $U \in \tau_{\mathrm{disc}}$ be an open neighbourhood of $a_0$. Since $a_0 \in U$ and $x_{k_m} = a_0$ for every $m$, all terms of the subsequence lie in $U$. Hence $x_{k_m} \to a_0$. Since the original sequence was arbitrary, every sequence in $X$ has a convergent subsequence, and so $(X, \tau_{\mathrm{disc}})$ is sequentially compact.
[/guided]
[/step]
[step:Construct a sequence with no convergent subsequence in an infinite discrete space]
We prove the contrapositive of the reverse implication. Assume that $X$ is infinite. Since we are working in the usual ZFC set-theoretic setting, choose a sequence $y: \mathbb{N} \to X$ such that $y_m \neq y_n$ whenever $m,n \in \mathbb{N}$ and $m \neq n$. Write $y_n := y(n)$.
Let $k: \mathbb{N} \to \mathbb{N}$ be any strictly increasing map, so that $y \circ k$ is an arbitrary subsequence of $y$. Suppose, for contradiction, that this subsequence converges to some point $p \in X$.
Since $\tau_{\mathrm{disc}} = \mathcal{P}(X)$, the singleton $\{p\}$ is open in $X$. By convergence of $y \circ k$ to $p$, there exists $M \in \mathbb{N}$ such that $y_{k_m} \in \{p\}$ for every $m \ge M$. Hence $y_{k_M} = p$ and $y_{k_{M+1}} = p$. But $k_M \neq k_{M+1}$ because $k$ is strictly increasing, and the terms of $y$ are pairwise distinct. This contradiction shows that no subsequence of $y$ converges.
Thus $X$ admits a sequence with no convergent subsequence, so $(X, \tau_{\mathrm{disc}})$ is not sequentially compact. Therefore, if $(X, \tau_{\mathrm{disc}})$ is sequentially compact, then $X$ must be finite.
[/step]
[step:Combine the two implications]
We have shown that every finite discrete space is sequentially compact. We have also shown that every infinite discrete space fails to be sequentially compact. Hence $(X, \tau_{\mathrm{disc}})$ is sequentially compact if and only if $X$ is finite.
[/step]
