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I like the following geometric argument.

Lemma. For any convex compact set $K\subset \mathbb{R}^2$ with non-empty interior there exists a point $x$ on the boundary such that the support line in $x$ is unique.

Proof. Let $p$ be interior point, and $x$ the closest to $p$ point of the boundary. Since the circle centered in $p$ and passing through $x$ is contained in $K$, the only support line in $x$ is perpendicular to the segment $px$.

Now assume that the points of $[0,1]$ are enumerated as $r_1,r_2,\dots$. Consider the convex function $f(x)=\sum_n 2^{-n}\max((r_n-1)x,r_n(x-1))$. Consider the set $K$ bounded by the graphs of $f$ and $-f$: $K=\{(x,y):f(x)\leqslant y\leqslant -f(x),x\in[0,1]\}$. It is easy to see that it does not satisfy the conclusion of Lemma, since the left derivative of $f$ is strictly smaller than the right derivative at any interior point of the segment $[0,1]$, and the derivative at endpoints is finite.

I like the following geometric argument.

Lemma. For any convex compact set $K\subset \mathbb{R}^2$ with non-empty interior there exists a point $x$ on the boundary such that the support line in $x$ is unique.

Proof. Let $p$ be interior point, and $x$ the closest to $p$ point of the boundary. Since the circle centered in $p$ and passing through $x$ is contained in $K$, the onliest support line in $x$ is perpendicular to the segment $px$.

Now assume that the points of $[0,1]$ are enumerated as $r_1,r_2,\dots$. Consider the convex function $f(x)=\sum_n 2^{-n}\max((r_n-1)x,r_n(x-1))$. Consider the set $K$ bounded by the graphs of $f$ and $-f$: $K=\{(x,y):f(x)\leqslant y\leqslant -f(x),x\in[0,1]\}$. It is easy to see that it does not satisfy the conclusion of Lemma, since the left derivative of $f$ is strictly smaller than the right derivative at any interior point of the segment $[0,1]$, and the derivative at endpoints is finite.

I like the following geometric argument.

Lemma. For any convex compact set $K\subset \mathbb{R}^2$ with non-empty interior there exists a point $x$ on the boundary such that the support line in $x$ is unique.

Proof. Let $p$ be interior point, and $x$ the closest to $p$ point of the boundary. Since the circle centered in $p$ and passing through $x$ is contained in $K$, the onliest support line in $x$ is perpendicular to the segment $px$.

Now assume that the points of $[0,1]$ are enumerated as $r_1,r_2,\dots$. Consider the convex function $f(x)=\sum_n 2^{-n}\max((r_n-1)x,r_n(x-1))$. Consider the set $K$ bounded by the graphs of $f$ and $-f$: $K=\{(x,y):f(x)\leqslant y\leqslant -f(x),x\in[0,1]\}$. It is easy to see that it does not satisfy the conclusion of Lemma, since the left derivative of $f$ is strictcly smaller than the right derivative at any interior point of the segment $[0,1]$, and the derivative at endpoints is finite.

I like the following geometric argument.

Lemma. For any convex compact set $K\subset \mathbb{R}^2$ with non-empty interior there exists a point $x$ on the boundary such that the support line in $x$ is unique.

Proof. Let $p$ be interior point, and $x$ the closest to $p$ point of the boundary. Since the circle centered in $p$ and passing through $x$ is contained in $K$, the only support line in $x$ is perpendicular to the segment $px$.

Now assume that the points of $[0,1]$ are enumerated as $r_1,r_2,\dots$. Consider the convex function $f(x)=\sum_n 2^{-n}\max((r_n-1)x,r_n(x-1))$. Consider the set $K$ bounded by the graphs of $f$ and $-f$: $K=\{(x,y):f(x)\leqslant y\leqslant -f(x),x\in[0,1]\}$. It is easy to see that it does not satisfy the conclusion of Lemma, since the left derivative of $f$ is strictly smaller than the right derivative at any interior point of the segment $[0,1]$, and the derivative at endpoints is finite.

I like the following geometric argument.

Lemma. For any convex compact set $K\subset \mathbb{R}^2$ with non-empty interior there exists a point $x$ on the boundary such that the support line in $x$ is unique.

Proof. Let $p$ be interior point, and $x$ the closest to $p$ point of the boundary. Since the circle centered in $p$ and passing through $x$ is contained in $K$, the onliest support line in $x$ is perpendicular to the segment $px$.

Now assume that the points of $[0,1]$ are enumerated as $a_1,a_2,\dots$. Consider the convex function $f(x)=\sum_n 2^{-n}\max((r_n-1)x,r_n(x-1))$. Consider the set $K$ bounded by the graphs of $f$ and $-f$: $K=\{(x,y):f(x)\leqslant y\leqslant -f(x),x\in[0,1]\}$. It is easy to see that it does not satisfy the conclusion of Lemma, since the left derivative of $f$ is strictcly smaller than the right derivative at any interior point of the segment $[0,1]$, and the derivative at endpoints is finite.

I like the following geometric argument.

Lemma. For any convex compact set $K\subset \mathbb{R}^2$ with non-empty interior there exists a point $x$ on the boundary such that the support line in $x$ is unique.

Proof. Let $p$ be interior point, and $x$ the closest to $p$ point of the boundary. Since the circle centered in $p$ and passing through $x$ is contained in $K$, the onliest support line in $x$ is perpendicular to the segment $px$.

Now assume that the points of $[0,1]$ are enumerated as $r_1,r_2,\dots$. Consider the convex function $f(x)=\sum_n 2^{-n}\max((r_n-1)x,r_n(x-1))$. Consider the set $K$ bounded by the graphs of $f$ and $-f$: $K=\{(x,y):f(x)\leqslant y\leqslant -f(x),x\in[0,1]\}$. It is easy to see that it does not satisfy the conclusion of Lemma, since the left derivative of $f$ is strictcly smaller than the right derivative at any interior point of the segment $[0,1]$, and the derivative at endpoints is finite.