Let $B(r):=\{x\in\mathbb{R}^n: \| x\| \leq r\}$ be a closed ball in $\mathbb{R}^n$ with respect to the Euclidean norm on $\mathbb{R}^n$ and let $p\in\partial B(r)$. I want to show that $B(r)\setminus \{p\}$ is convex only using the definition of convexity (i.e. for any two points in the set, the whole segment connecting the two points lies also in the set).
My attempt: Let $x,y\in B(r)\setminus\{p\}$ and $\lambda\in[0,1]$. As balls are convex, we have that $(1-\lambda)x+\lambda y\in B(r)$. I then want to show $(1-\lambda)x+\lambda y\neq p$. Assume the contrary, i.e. $(1-\lambda)x + \lambda y = p$.
We can then distinguish three cases: If $\lambda = 0$, we have $x = p$, a contradiction to $x\in B(r)\setminus \{p\}$. If $\lambda = 1$, we have $y=p$ which is a contradiction to $y\in B(r)\setminus\{p\}$. In the last case, let $0 < \lambda < 1$. We then estimate \begin{equation} r = \|p\| = \|(1-\lambda)x + \lambda y\| \leq (1-\lambda) \|x\| + \lambda \|y\|< \|x\| + \|y\|. \end{equation} Can anyone help me to obtain a contradiction here?
You're almost there. So far you have $$r = \|(1-\lambda) x + \lambda y\| \overset{(i)}{\le} (1-\lambda) \|x\| + \lambda \|y\| \overset{(ii)}{\le} (1-\lambda) r + \lambda r = r.$$
To obtain a contradiction, we show that either (i) or (ii) must be a strict inequality.
(ii) attains equality if and only if $\|x\|=\|y\|=r$. If this fails to hold, then (ii) is a strict inequality and we are done.
Thus, to handle the remaining case, assume $\|x\|=\|y\|=r$ and assume (i) is an equality. We will show a contradiction.
In Euclidean space, the triangle inequality (i) attains equality if and only if $x$ and $y$ are parallel (specifically, when $x \cdot y = \|x\| \|y\|$, see the application of Cauchy-Schwarz in the proof of the triangle inequality). But since $\|x\|=\|y\|$ this requires that $x=y$. But then $p=(1-\lambda)x + \lambda y = x$ which contradicts $x \in B \setminus \{p\}$.