Proving that $\{\vec{x}\in\mathbb{R}:\lVert \vec{x}\rVert\ge 1 \}$ is closed To prove that  $\{\vec{x}\in\mathbb{R}:\lVert \vec{x} \rVert\ge 1 \}$ is closed, I need to use the following definition of a closed set:

Let $V$ be a normed vector space. A set $S\subseteq V$ is closed iff it contains all its limit points.

In my understanding, I need to show that if $\exists \vec{x}\in V$ such that a sequence $\{\vec{x}_n\}$ contained entierely in $S$ converges to $\vec{x}$, and $\vec{x}\not\in S$, then $\exists \varepsilon>0$ such that $n>N_\varepsilon >0$ implies $x_n\not\in S$, a contradiction. However, I have no idea  how to do this.
Would appreciate some help.
 A: It follows from the reverse triangle inequality 
$$|\;||x||-||y||\;|\leq ||x-y||$$
that the norm is continuous, hence if $\{x_n\}$ is a sequence with $||x_n||\geq 1$ for all $n$ and $x_n\to x$, then
$$ ||x||=\lim_{n\to\infty}||x_n||\geq 1$$
A: I was in a similar trouble yesterday completing a very similar proof for the closure of a set under a normed space.
The unique that you need to focus is in prove that if for all $n$ we have that $\|x_n\|\ge 1$ and if $(x_n)\to x$ then $\|x\|\ge 1$.
This comes from the theorem that says that for converging sequences $(x_n)$ and $(y_n)$
$$y_n<x_n,\forall n\in\Bbb N\implies \lim y_n\le \lim x_n$$
and this is obviously true too if we have that $y_n\le x_n$ (we dont need the strict inequality). Then if we set as a constant sequence $y_n=w$ such that $\|w\|=1$ for all $n$ then $\lim \|y_n\|=1\le \lim\|x_n\|$.

Another way to prove it is characterizing a convergence sequence under it open balls, i.e. if we have that for some $(x_n)\to x$
$$\forall\epsilon>0,\exists N\in\Bbb N:\|x_n-x\|<\epsilon,\forall n\ge N$$
then we can write too
$$\forall\epsilon>0,\exists N\in\Bbb N:x_n\in \Bbb B(x,\epsilon),\forall n\ge N$$
where $\Bbb B(x,\epsilon):=\{y\in V:\|x-y\|<\epsilon\}$. Then for any $\epsilon>0$ almost all points of the sequence are contained in $\Bbb B(x,\epsilon)$.
Now observe if the elements of $(\|x_n\|)$ belong almost all (or not) to some ball $\Bbb B(\|x\|,\epsilon)$ such that $\|x\|<1$.
P.S.: here almost all mean all but finitely many.
