Proving a set is open. 
Let $E,||\cdot||$ be a finite dimensional normed space over $\mathbb R$.
Let $U$ be an open subset of $E$ and $a\in U$
Let $A=\{x\in E \;|\; \forall t\in [0,1], (1-t)a+tx\in U\}$
Prove that $A$ is open.

Intuitively, $A$ is the set of points $x$ such that the line segment from $a$ to $x$ lies inside $U$.
Pick some $x\in A$. One needs to find some $r>0$ such that the open ball centered at $x$ with radius $r$ is a subset of $A$. Since $a\in U$ and $x\in U$, there are some $r_1$ and $r_2$ such that $B(r_1,a)\subset U$ and  $B(r_2,a)\subset U$. Setting $r:=\min(r_1,r_2)$ seems plausible, but I can't prove that it fits the bill.
I tried contradiction as well, to no avail. This yields the existence of $x_n$ such that $x_n\to x$ and $x_n \notin A$. That seems impossible, but I can't reach any contradiction.

If $x\in A$ then $x\in U$: Just take $t=1$ in the definition of $A$.  
Fix $x\in A$ and let $X$ denote the closed line segment from $a$ to $x$. As you noted, $X\subset U$.
Since $U$ is open, for each $y \in X$ there is $\varepsilon_y>0$ such that 
$B(\varepsilon_y,y)\subseteq U$.
Since $X$ is compact there is $\lambda_x>0$ such that $B(\lambda_x,y)\subseteq U$ for all $y \in X$.
(Use the Lebesgue number for the relative open cover of $X$ by the $B(\varepsilon_y,y)$, $y\in X$.
Formally if $\delta$ is the Lebesgue number then we may pick $\lambda_x<\delta/2$.
Then, for any $y\in X$ we have $\mathrm{diam}\, B(\lambda_x,y)<\delta$, hence
$B(\lambda_x,y)\subseteq B(\varepsilon_v,v)\subseteq U$ for some $v\in X$.)
Now take any $z$ in $B(\lambda_x,x)$. Clearly the line segment $Z$ from $a$ to $z$ lies in $B(\lambda_x,X):=\cup\{B(\lambda_x,y):y \in X)\} \subseteq U$.
It follows that $z\in A$, hence $B(\lambda_x,x) \subseteq A$.
This completes the proof that $A$ is open. 

