Proof of Compact Subsets of a Metric Space are Closed and Borel

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is used.

Claim 1. By Sequential Characterization of Closed Subsets of a Metric Space, the set $K$ is closed in $(X,\mathcal{T}_{d})$ as soon as the following holds: for every sequence $(x_{p})_{p\in\mathbb{N}}$ in $X$ with $x_{p}\in K$ for every $p\in\mathbb{N}$, and every $x\in X$ to which $(x_{p})_{p\in\mathbb{N}}$ converges in $(X,d)$, one has $x\in K$. Let such a sequence $(x_{p})_{p\in\mathbb{N}}$ and such a point $x$ be given.

By A Compact Subset of a Metric Space is Sequentially Compact, $K$ is sequentially compact in $(X,d)$. Applying that definition to the sequence $(x_{p})_{p\in\mathbb{N}}$, whose terms lie in $K$, there are a point $y\in K$ and a strictly increasing sequence $(p_{k})_{k\in\mathbb{N}}$ in $\mathbb{N}$ such that the subsequence $(x_{p_{k}})_{k\in\mathbb{N}}$ converges to $y$ in $(X,d)$.

By A Subsequence of a Convergent Sequence Has the Same Limit, applied to the convergent sequence $(x_{p})_{p\in\mathbb{N}}$ with limit $x$ and to the same strictly increasing sequence $(p_{k})_{k\in\mathbb{N}}$, the subsequence $(x_{p_{k}})_{k\in\mathbb{N}}$ converges to $x$ in $(X,d)$. That subsequence therefore converges both to $x$ and to $y$, so $x=y$ by Uniqueness of Limits in a Metric Space. Since $y\in K$, we get $x\in K$, which is what was to be shown.

Claim 2. By claim 1, $K$ is a closed subset of $(X,\mathcal{T}_{d})$, so $K\in\mathcal{B}(X)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space, applied to the metric space $(X,d)$. By Sigma-Algebra and Measurable Space a $\sigma$-algebra on $X$ contains the complement in $X$ of each of its members, and $\mathcal{B}(X)$ is a $\sigma$-algebra on $X$ by Borel Sigma-Algebra of a Metric Space; hence $X\setminus K\in\mathcal{B}(X)$.