I'm trying to prove that the unit sphere $$S=\{x\in X \:\:|\: ||x||=1\}$$ is compact in any normed vector space $X$.
If $X$ was $\mathbb{R}^n$, then I could use Heine-Borel and the result would be trivial. Can't I use the fact that $X$ is isomorphic to $\mathbb{R}^n$? If so, how?
I can't use equivalence of norms to prove this since I am using this theorem to prove the equivalence of norms but I can use $||x||=||x||_1$.
PS: There are several posts here trying to prove this result but none using the isomorphism of $X$ to $\mathbb{R}^n$ and explaining this step convincingly.
In a finite-dimensional normed linear space $X$ over $\mathbb R$ with basis $b_1, \ldots, b_n$, the map $f: (c_1, \ldots, c_n) \to c_1 b_1 + \ldots + c_n b_n $ from $\mathbb R^n$ to $X$ is continuous (by continuity of scalar multiplication and addition in $X$), one-to-one and onto (by definition of basis). The closed unit ball $B$ of $\mathbb R^n$ is compact, so $f(B)$ is compact. So is $f(rB) = r f(B)$ for any $r > 0$, where $rB = \{r x: x \in B\}$.
Since $x \mapsto \|f(x)\|$ is a continuous real-valued function on the unit sphere of $\mathbb R^n$ (which is a compact set), it attains a minimum value $L$ there, and $L > 0$ because $f$ is one-to-one. Thus $\|f(x)\| \ge L > 0$ for all $x \in \mathbb R^n$ with $\|x\| = 1$, and by homogeneity $\|f(x)\| \ge L \|x\|$ for all $x \in \mathbb R^n$. In particular, your unit sphere $S$ of $X$ is contained in $f(rB)$ for $r = 1/L$.
Now $S$ is a closed (because the norm is continuous) subset of a compact set, therefore it is compact.