Under a finite dimensional norm-induced metric space, is every bounded sequence has a converging subsequence? What I know is that every bounded sequence in $\Bbb R^n$ has a converging sub sequence regardless of the chosen norm, since this is true under the Euclidean norm, and all norms on $\Bbb R^n$ are equivalent. My question is 

Is every bounded sequence has a converging sub sequence for any norm-induced metric space? 

If not, a counter-example is appreciated. Thanks!

I assume that by finite dimensional norm-induced metric space you mean a vector space in the first place because otherwise the concept of a norm or dimension doesn't make any sense.
If $(V, \|\cdot\|)$ is any $n$-dimensional normed $\mathbb F$-Vector space, where $\mathbb F\in\{\mathbb R,\mathbb C\}$ and $\{b_1,...,b_n\}$ is a basis of $V$, then the unique linear map $\varphi$ that maps $b_i$ to $e_i$ where $\{e_1,...,e_n\}$ is the standard base of $\mathbb F^n$ is an isomorphism. It is easily checked that $|\cdot|$ defined by $|x| := \|\varphi^{-1}(x)\|$ defines a norm on $\mathbb F^n$. For this norm it is true that each bounded sequence has a convergent subsequence. If $(x_n)_{n\in\mathbb N}$ is a bounded sequence in $V$, then $(\varphi(x_n))_{n\in\mathbb N}$ is bounded in $(\mathbb F^n, |\cdot|)$, so it has a convergent subsequence $(\varphi(x_{n_k}))_k$, but then $(x_{n_k})_k$ must converge in $V$ because $\varphi^{-1}$ is continuous.
