Let $X$ be a normed linear space and let $V\subset X$ be a linear subspace. Prove that if $V$ is open, then $X=V$.
We need to prove that $X\subset V$. Let there be a sequence $x_n\in X$ such that $x_n\to x$ with $x\in V$. There exists $N\in\mathbb{Z}$ such that for all $n\geq N$ $x_n\in V$. Hence there is an open ball $B(x_n;\epsilon)\subset V$ with $\epsilon >0$, around $x_n$.
Here is where I need help. Intuitively I know that $$x_n\in B(x_n;\epsilon) \implies x_{n-1}\in B(x_n;\epsilon)$$
Can I just make $x_n$ arbitrarily close to each other? Is there a more rigorous approach to tackle this last part? Maybe using the fact that $V$ is closed under addition and scalar multiplication of vectors?