Let $\left<M,||.|| \right>$ be a normed vector space, show that $G = \{x \in M \mid \|x\| \leq 1\}$ is closed.
My strategy is to show $G^c$ is open.
Let $\rho$ be the metric of $M$ and $G^c=\left\{x \in M \ \middle| \ \ \Vert x \Vert >1 \right\}$
Let $r = \Vert{x}\Vert -1 \gt 0$
Let $y \in B_r(x)$ then $\rho(x,y) \lt r \iff \rho(x,y) \lt \Vert x \Vert - 1$
By triangle inequality of metrics,
$\rho(x,0) \leq \rho(x,y) + \rho(y,0)$
$\implies \Vert x \Vert \lt \Vert x \Vert - 1 + \rho(0,y)$
$\iff 1 \lt \rho(0,y)$
$\iff \Vert y \Vert \gt 1$
$\implies y \in G^c \implies B_r(x) \subset G^c \implies G^c$ is open $\implies G$ is closed $\ \ \ \Box$
The basis of my proof is that norm of vector $x$ is the metric between $x$ and the $0$ vector. Is this okay?