Prove that this is a norm

Consider $x,y\ne 0$ and define $\tilde{x}=\frac{1}{N(x)}x$, $\tilde{y}=\frac{1}{N(y)}y$. Then, by property 3., $N(\tilde{x})=1=N(\tilde{y})$. So $\tilde{x},\tilde{y}$ belong to $B$. Let's use convexity then: $$ z=\frac{N(x)}{N(x)+N(y)}\tilde{x} + \frac{N(y)}{N(x)+N(y)}\tilde{y} \in B.$$ This yields $N(z)\leq 1$ which is equivalent to $$N(N(x)\tilde{x}+N(y)\tilde{y})\leq N(x)+N(y)$$by property 3. again.

Now, since $N(x)\tilde{x}=x$ and $N(y)\tilde{y}$, by definition, we're done.