Equivalence of norms on a finite dimension space without using compactness of the unit ball I have seen and understand the proof of equivalence of norms when we assume the compactness of the unit ball. However, I have been asked
Assume that all finite-dimensional subspaces of a normed space are closed. Prove that all norms on a finite-dimensional normed space are equivalent without assuming the unit ball is compact. You may start by showing that all norms on a 1-dimensional normed space are equivalent. Then apply mathematical induction.
I am very lost on how to prove this. I can easily get the upper (or lower bound depnding on the norm used) but I am having a very hard time with the induction part of the argument.
 A: Consider any normed linear space $X$. If the kernel of a linear functional is closed then the functional is continuous. Proof: let $f$ be a non-zero functional and choose $y$ such that $f(y)=1$. If $f$ is not continuous then there exists $(x_n)$ such that $|f(x_n)| >n$ and $\|x\|_n=1$ for all $n$. But then $\frac {x_n} {f(x_n)}-y$  is a sequence in the kernel of $f$ converging to $-y$ and $f(-y) \neq 0$ leading to a  contradiction. 
Now your result follows immediately: if $X$ is finite dimensional  space and  every subspace is  assumed to be closed then every linear functional is continuous. Hence the identity map from $(X,\|.\|_1) \to (X,\|.\|_2)$ is continuous  for any two norms $\|.\|_1$  and $\|.\|_2$ proving that any two norms are equivalent. 
A: Just to point out, what the answer above actually does is to show that if $X$ is a finite-dimensional normed vector space, then any linear map $\phi\colon X \to Y$ is continuous: exactly as it points out for the identity map, the continuity for a general linear map $\phi$ follows by, say, picking a basis $\{f_1,\ldots, f_k\}$ for $\phi(X)\leq Y$ (the image of $X$ is finite-dimensional because $X$ is) and writing $\phi(v) = \sum_{i=1}^k \phi_i(v).f_i$, where the $\phi_i$ are linear functionals on $X$.
You can then use induction to prove the continuity of linear functionals on a finite-dimensional vector space: if $f\colon X \to \mathbb R$ is a linear functional, then $H=\ker(f)$ is a subspace of dimension $n-1$ where $n = \dim(X)$. Now it is enough to show that $H$ is complete, because a complete subspace of a metric space has to be closed. To see that $H$ is complete, pick a basis of $H$, $\{e_1,\ldots,e_{n-1}\}$ and let $\psi \colon \ell_2^n \to H$ (where $\ell_2^n$ denotes the $n$-dimensional Euclidean vector space/inner product space) be given by $\psi((a_1,\ldots,a_{n-1}) \mapsto \sum_{i=1}^{n-1} a_ie_i$. Now $\psi$ is an isomorphism of linear spaces, and by the induction hypothesis, $\psi$ is bounded. But since $\ell_2^n$ is complete and any bounded linear map is uniformly continuous, it follows that $H$ is complete as required!
[Just for completeness, the continuity of any linear map $f\colon X\to Y$ when $\dim(X)=1$ really is trivial: if we pick a vector $e \in X$ with $\|e\|=1$, then $\|f\| = |f(e)|$.]
