Let $E$ be a normed vector space and $F$ a sub-vector space
How to prove that $$\mathring{F}\neq\varnothing \implies F=E$$
I suppose that $\mathring{F}\neq\varnothing $ is the largest open subspace of $F$, that is $\exists r>0, B(x,r)\subset F$
but I don't know how to continue.
Let $y \in \text{int}(F)$. So, $$B(y, \varepsilon) = \lbrace x \in E; ||x - y|| < \varepsilon \rbrace \subset F (*)$$
Let $0 \neq z \in E$. Consider the vector $x \in E$, such that : $x = y + \dfrac{z}{2||z||}\varepsilon $.
Thus, $||x - y|| = \dfrac{||z||}{||z||} \dfrac{\varepsilon}{2} = \dfrac{\varepsilon}{2} < \varepsilon$.
By (*), we have $x \in B(y, \varepsilon)$ $\Rightarrow x \in F$, therefore $E \subset F$. So, $E = F$.
The Reciprocal is immediate.