Let $A\subset \Bbb R^n$ be a bounded, symmetric, convex, open set.
Let $\|x\|=\inf\{k>0 : x/k\in A\}$.
$\bullet$ $\|0\|=0$
$0\in A$ because $A$ is symmetric and convex (because $0=x/2-x/2\in A$). Thus $0/k \in A$ for every $k>0$ implying that $\|0\|=0$.
$\bullet$ $\|t x\|=|t|\|x\|\quad \forall t\in \Bbb R$
Let $t>0$, then it holds
$$ \|tx\| = \inf\Big\{k>0 : \frac{tx}{k}\in A\Big\}= t\inf\Big\{\frac{l}{t}>0 : \frac{x}{\frac{l}{t}}\in A\Big\}=t\|x\|.$$
Now, as $A$ is symmetric, we have $x/k\in A$ if and only if $-x/k\in A$ which implies that $\|x\|=\|-x\|$. Hence, if $t<0$, then $$\|tx\|=\||t|(-x)\|=|t|\|-x\|=|t|\|x\|.$$
$\bullet$ $\|x+y\|\leq \|x\| + \|y\|$
Let $x,y\in \Bbb R^n$, then for every $k,l>0$ such that $x/k\in A$ and $y/l\in A$ it holds
$$ \frac{k}{k+l}\frac{x}{k}+\frac{l}{k+l}\frac{y}{l}=\frac{x+y}{k+l}\in A$$
by convexity of $A$. It follows that
\begin{align*}
\|x+y\| &= \inf\Big\{m>0 : \frac{x+y}{m}\in A\Big\}\\
&\leq \inf\Big\{k>0 : \frac{x}{k}\in A\Big\}+\inf\Big\{l>0 : \frac{y}{l}\in A\Big\}\\ &=\|x\|+\|y\|.\end{align*}
$\bullet$ $\|x\|=0 \ \implies \ x=0$
If $\|x\|=0$, then $x/k\in A$ for every $k>0$. But $A$ is bounded, and so, if $x\neq 0$, we get a contradiction by letting $k\to 0$. Hence, we must have $x=0$.
Combining the first and last point we get $\|x\|=0$ if and only if $x=0$ which shows that $\|\cdot \|$ is a norm on $\Bbb R^n$.
Remark: The reverse direction of the exercise can be shown. Indeed, if $\|\cdot\|'$ is a norm on $\Bbb R^n$, then $A'=\{x\mid \|x\|'< 1\}$ is a bounded, symmetric, convex, open set.
For the last part, let $B=\{x\in \Bbb R^n\mid \|x\|<1\}$ and $z\in A$. Then, we have $z=\frac{z}{1}\in A$ implying that $\|z\| <1$, i.e. $z\in B$. It follows that $A\subset B$.
Triangle's inequality in the definition of a norm is as follows:
$$\forall x,y\in X,\ \|x+y\|\le \|x\|+\|y\| $$
Now, the above inequality follows the convexity of the unit closed ball. Let $\lambda\in [0,1]$ and $\mathbf{x},\mathbf{y}\in \mathbb{B}[0,1]\implies \|\mathbf{x}\|\le 1 $ and $\|\mathbf{y}\|\le 1$. Now consider,
\begin{align*}
\|\lambda \mathbf{x}+(1-\lambda) \mathbf{y}\| & \le \|\lambda \mathbf{x}\|+\|(1-\lambda)\mathbf{y}\|\\
& = \lambda \|\mathbf{x}\|+(1-\lambda)\|\mathbf{y}\|\\
& \le \lambda + (1-\lambda)=1
\end{align*}
Hence, $ \lambda \mathbf{x}+(1-\lambda) \mathbf{y}\in \mathbb{B}[0,1]. $
For the converse see,
$$ \frac{\mathbf{x}+\mathbf{y}}{\|\mathbf{x}\|+\|\mathbf{y}\|}=\frac{\|\mathbf{x}\|}{\|\mathbf{x}\|+\|\mathbf{y}\|}\frac{\mathbf{x}}{\|\mathbf{x}\|}+\frac{\|\mathbf{y}\|}{\|\mathbf{x}\|+\|\mathbf{y}\|}\frac{\mathbf{y}}{\|\mathbf{y}\|}\in \mathbb{B}[0,1] \implies \|\mathbf{x}+\mathbf{y}\|\le \|\mathbf{x}\|+\|\mathbf{y}\|$$
Best Answer
There's a few ways to proceed. Probably the easiest is to not worry about the unit ball, and just prove the triangle inequality directly (that proof is available everywhere, for example on Wikipedia, if you want to look), but you asked for a version via convexity, so I'll give it a go:
First, note that $\|af+(1-a)g\|_{L_p} \leq 1$ if and only if $\|af+(1-a)g\|_{L_p}^p \leq 1$, so we'll deal with the latter, since it gets rid of that awkward power of $1/p$.
At this point, things are relatively easy: first, we note that $x \mapsto x^p$ is convex on the non-negative real half-line (feel free to insert a proof of this here if you don't have one already), so (combining this with the triangle inequality for $|\cdot|$) we have $|af(x)+(1-a)g(x)|^p \leq \left(a|f(x)| + (1-a)|g(x)| \right)^p\leq a|f(x)|^p + (1-a)|g(x)|^p$.
Thus, by standard properties of integrals, the definition of $\|\cdot\|_{L_p}$, and the fact that $f$ and $g$ lie in the closed unit ball, we have
\begin{align*}\|af+(1-a)g\|_{L_p}^p &= \int |af(x)+(1-a)g(x)|^p\mathrm{d}x \\&\leq \int a|f(x)|^p+(1-a)|g(x)|^p\mathrm{d}x \\&= a\int|f(x)|^p\mathrm{d}x + (1-a)\int|g(x)|^p\mathrm{d}x \\&= a\|f\|_{L_p}^p + (1-a)\|g\|_{L_p}^p \\&\leq a + (1 -a) \\&= 1\end{align*}
Thus, $\|af+(1-a)g\|_{L_p} \leq 1$ also, so our open unit ball is convex, hence we have the triangle inequality.