Prove that the set $\{x_n:n\in\mathbb{N}\}\cup\{a\}$ is sequentially compact using sequences

Question

NOTE: This question has been asked more than once, but the other answers use open covers, not sequences, and do not interest me at the moment. I just want more refined mathematicians to look at my proof attempt, correct, improve, or criticize.

STATEMENT:
Let $(X,d)$ be a metric space and let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$ that converges to $a\in X$.
Prove that the set $\{x_n:n\in\mathbb{N}\}\cup\{a\}$ is compact.

ATTEMPT:
Let $a=x_\infty$ and $S=\{x_n:n\in\mathbb{N}\}\cup\{x_\infty\}$.

Case 1: $x_n$ has finitely many values. $$S \text{ is finite}\Longrightarrow S \text{ is compact}$$

Case 2: $x_n$ has infinitely many values.

Let $(y_k)_{k\in\mathbb{N}} \in S$ be an arbitrary sequence.
(Not necessarily a subsequence of $x_n$, can be a reordering)

Let $f:=y_k\to n$ and $g:=n \to k$.
($f$ maps the value of $(y_k)$ to the index of the corresponding $x_n$)
($g$ maps that index back to the index of $y_k$)

Let $y_{k_l}$ be a subsequence of $y_k$ such that: $$\cases{k_0=g(0)\\ k_l=g\bigg(\text{min}\big\{ f\big((y_k)_{k>k_{l-1}}\big)\big\}\bigg)}$$

(I am sorry about this horrible behemoth. What I want it to do is:)

1. Start at the index of $y_k=x_0$.
2. Map all the succeeding terms of $y_k$ to their $n$-value.
3. Take the minimum $n$.
4. Map that minimum to its $k$-value.

Example:
$(y_k)=(x_9,x_2,x_0,x_5,x_1,x_7,x_3,...)$
$(y_{k_l})=(x_0,x_1,x_3,...)$

The resulting subsequence $(y_{k_l})$ of $(y_k)$ is also a subsequence of $(x_n)$.

Since $x_n \to x_\infty$ then $y_{k_l} \to x_\infty$.

Therefore $\forall y_k \in S, \exists y_{k_l} \to x_\infty \in S$.

Therefore $S$ is compact.

I am well-aware how inelegant this proof is.
I am not even sure if it is correctly written or expressed.
Help, please.

Answer 1

I believe that your idea will work, but I agree that the notation and write-up could use some improvement. Since you have an algorithm in mind, let’s use a proof with an algorithm. Let’s consider the set $$N_0=\{n\in \mathbb{N}:x_n\in (y_k)\}.$$ Then $N_0$ is non-empty and a subset of the natural numbers, so it has a minimum, say $n_0$. Let us set $z_0=x_{n_0}$. Now construct $$N_1=\{n\in \mathbb{N}:x_n\in (y_k), n>n_0\}=N_0\cap \{n\in\mathbb{N}:n>n_0\}.$$ If $N_1$ is empty, then $(y_k)$ was constant, so it converges. Thus, we may assume $N_1\neq\emptyset$. Take the minimum, $n_1$. Set $z_1=x_{n_1}$. Inductively construct this sequence and then you can “observe” (as you did above) that it is a subsequence of $(y_k)$ and a subsequence of $(x_k)$, so it will converge.

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For good measure, I will write out one more step: Let $$N_2=N_{1}\cap \{n\in \mathbb{N}:n>n_{1}\}.$$ If $N_2$ is empty, then $(y_k)$ contains only $2$ distinct values, and we may find a subsequence of it that converges (the idea is that at least one of the values must be repeated infinitely often, and there are only finitely-many values, so we just pick the constant subsequence whose value is the one that is repeated infinitely-often). Otherwise, we have a minimum for $N_2$ say $n_2$ and set $z_2=x_{n_2}$. At each stage, this algorithm either terminates with us choosing a constant subsequence, or it does not and we construct a sequence $(z_i)=(x_{n_i})$ which converges because $(x_n)$ does.