If $\varphi \in E'$ and $A$ is convex and open then $\varphi (A)$ is an open interval 
Let $E$  be a real normed space and $\varphi \in E'$, $\varphi \neq 0$.
  Suppose that $A \subset E$ is an open convex not empty subset.
  Show that $\varphi(A)$ is an open interval.

Since $A$ is connected and $\varphi$ is continuous,  $\varphi(A)$ is an interval.
How could I show that $\varphi(A)$  is open? 
I tried to apply the geometric form of Hahn Banach theorem but it didn't work.
I appreciate if you could give some hints. Thanks.

Since $\varphi(A)$ is an $\mathbb{R}$-interval, it is of the form $(a,b)$, $[a,b]$, $(a,b]$, $[a,b)$, with $-\infty\le a,b\le\infty$. Supose $\varphi(A)$ is not open. Then, WLG we can assume $\varphi(A)=[a,b)$. Then, there is $x\in A$ such that $\varphi(x)=a$. 
Since $A$ is open, there is $r>0$ such that if $\|x-y\|<r$ we have $y\in A$. Now, if $\|z\|<r$ we have $\|x-(x+z)\|<r$ and thus $a\le \varphi(x+z)=a+\varphi(z)$. Therefore $\varphi(z)\ge 0$ for all $z$ with $\|z\|<r$.
Fix $z_0$, $\|z_0\|<r$. Then $\|-z_0\|<r$. We have $0\le\varphi(-z_0)=-\varphi(z_0)\le 0$. Thus, $\varphi(z)=0$ for all $z$ with $\|z\|<r$.
But the last implies, by linearity of $\varphi$, that $\varphi=0$ (cause every element of the space is a scalar multiple of an element of norm $<r$). This contradiction shows $\varphi(A)\neq [a,b)$ for all $a$.
