= Solution
A <norm> $p$ is nonnegative with $p(x)=0$ exactly when $x=0$, satisfies $p(ax)=|a|p(x)$, and obeys $p(x+y)\leq p(x)+p(y)$. The <Euclidean norm> has positivity and absolute homogeneity immediately. For the <triangle inequality>, positivity of $\|x-ty\|_2^2$ for all real $t$ gives the <Cauchy-Schwarz inequality> $|x\cdot y|\leq\|x\|_2\|y\|_2$ (with $y=0$ immediate). Hence
$$
\|x+y\|_2^2\leq\|x\|_2^2+2\|x\|_2\|y\|_2+\|y\|_2^2,
$$
which proves the <triangle inequality> after taking square roots.
For the prescribed set, use the <Minkowski functional>
$$
\boxed{p_U(x)=\inf\{t>0:x\in tU\}.}
$$
Since $U$ is open and contains zero, some <Euclidean ball> of radius $r>0$ lies inside it. Since it is bounded, it lies inside a ball of radius $R$. These inclusions imply $\|x\|_2/R\leq p_U(x)\leq\|x\|_2/r$, so the functional is finite and positive for nonzero $x$; it is zero at zero. Symmetry gives $p_U(-x)=p_U(x)$, and substitution in the infimum gives $p_U(ax)=a p_U(x)$ for $a>0$, hence absolute homogeneity for all real scalars.
Convexity and $0\in U$ imply $sU\subseteq U$ for $0\leq s\leq1$. Consequently, for $a>p_U(x)$ and $b>p_U(y)$, one has $x/a,y/b\in U$. Convexity then gives
$$
\frac{x+y}{a+b}=\frac a{a+b}\frac xa+\frac b{a+b}\frac yb\in U,
$$
so $p_U(x+y)\leq a+b$. Letting $a\downarrow p_U(x)$ and $b\downarrow p_U(y)$ proves the <triangle inequality>.
Finally, if $p_U(x)<1$, there is a $t<1$ with $x\in tU\subseteq U$. Conversely, if $x\in U$ and $x\ne0$, openness permits $(1+\delta)x\in U$ for some $\delta>0$, so $p_U(x)\leq1/(1+\delta)<1$; zero is immediate. Thus \b[$\boxed{U=\{x:p_U(x)<1\}}$], completing the <norm from a bounded symmetric convex neighbourhood> construction.
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