Proof of Heine-Borel Theorem in $\mathbb{R}^n$

Claim 1 implies claim 2. Suppose $A$ is compact in $(\mathbb{R}^n,\mathcal{T}_{d_E})$. Then $A$ is closed in $(\mathbb{R}^n,\mathcal{T}_{d_E})$ by Compact Subset of $\mathbb{R}^n$ is Closed, and $A$ is bounded in $(\mathbb{R}^n,d_E)$ by Compact Subset of $\mathbb{R}^n$ is Bounded.

Claim 2 implies claim 1. Suppose $A$ is closed in $(\mathbb{R}^n,\mathcal{T}_{d_E})$ and bounded in $(\mathbb{R}^n,d_E)$. We show that $A$ is sequentially compact in $(\mathbb{R}^n,d_E)$.

Let $(x_m)_{m\in\mathbb{N}}$ be a sequence in $\mathbb{R}^n$ with $x_m\in A$ for every $m\in\mathbb{N}$. Since $A$ is bounded, Bolzano-Weierstrass Theorem in Euclidean Space provides a point $\ell\in\mathbb{R}^n$ 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 $\ell$ in $(\mathbb{R}^n,d_E)$.

The subsequence $(x_{p_k})_{k\in\mathbb{N}}$ is itself a sequence in $\mathbb{R}^n$, and each of its terms $x_{p_k}$ lies in $A$ because every term of $(x_m)_{m\in\mathbb{N}}$ does. Since $A$ is closed and this sequence converges to $\ell$ in $(\mathbb{R}^n,d_E)$, the sequential characterization of closed subsets, Sequential Characterization of Closed Subsets of a Metric Space, gives $\ell\in A$.

Thus every sequence in $\mathbb{R}^n$ with all terms in $A$ has a subsequence converging to a point of $A$, so $A$ is sequentially compact in $(\mathbb{R}^n,d_E)$. By Compactness and Sequential Compactness Agree for Subsets of a Metric Space the set $A$ is compact in $(\mathbb{R}^n,\mathcal{T}_{d_E})$.