Proof of Compact Subset of $\mathbb{R}^n$ is Closed

By Compactness and Sequential Compactness Agree for Subsets of a Metric Space, applied to the metric space $(\mathbb{R}^n,d_E)$ and the subset $A$, compactness of $A$ in $(\mathbb{R}^n,\mathcal{T}_{d_E})$ gives that $A$ is sequentially compact in $(\mathbb{R}^n,d_E)$.

We verify the sequential condition appearing in Sequential Characterization of Closed Subsets of a Metric Space. Let $(x_m)_{m\in\mathbb{N}}$ be a sequence in $\mathbb{R}^n$ with $x_m\in A$ for every $m\in\mathbb{N}$, and let $x\in\mathbb{R}^n$ be such that $(x_m)_{m\in\mathbb{N}}$ converges to $x$ in $(\mathbb{R}^n,d_E)$.

By sequential compactness there are a point $y\in A$ and a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$ such that the subsequence $(x_{n_k})_{k\in\mathbb{N}}$ converges to $y$ in $(\mathbb{R}^n,d_E)$. Since $(x_m)_{m\in\mathbb{N}}$ converges to $x$, A Subsequence of a Convergent Sequence Has the Same Limit shows that the same subsequence $(x_{n_k})_{k\in\mathbb{N}}$ converges to $x$ in $(\mathbb{R}^n,d_E)$. By Uniqueness of Limits in a Metric Space we conclude $x=y$, and therefore $x\in A$.

Thus every sequence with all terms in $A$ that converges in $(\mathbb{R}^n,d_E)$ has its limit in $A$, so Sequential Characterization of Closed Subsets of a Metric Space shows that $A$ is closed in $(\mathbb{R}^n,\mathcal{T}_{d_E})$.