$X$ be a normed space and assume that $E \subset X$ such that $\operatorname{int}(E) \neq\varnothing$ Let $X$ be a normed space and assume that $E \subset X$ such that $\operatorname{int}(E) \neq \varnothing$ then show that $E$ spans $X$.

I am trying it in a  following  way....
Let be the norm $\|\cdot\|:X \rightarrow \mathbb{R}^{+} \cup \{0\}$ , i am trying to write 
$$X = \bigcup_{\alpha \in \mathbb{R}^{+} \cup \{0\}}\|\cdot\|^{-1}(\alpha)$$
and then somehow embedding the a ball a interior point of $E$ such that ball entirely lies inside this union ...these are the thin gs comming in mind plesae help find the solution this problem as soon as possible.i don't know whether this question easy or tricky...

Continue the proof of Davide Giruado.
As $B(x_0,r) \subset E $ then  $B(0,r)=B(x_0,r)-x_0 \subset E $
Let $x$ be the arbitrary element of $X$, Choose $n$ such that $\left \| \frac{x}{n} \right \|< r$. So $\frac{x}{n} \in E$ which gives $x\in E$.
Consequently $X\subset E$ which deduces that $X=E$ as claimed.
