Understanding a "Trivial" Result In Optimal Control Theory

Sorry, I'm not an expert. However, I try to give you my insight with the hope that it helps.
In order to check whether $0$ is a boundary point of $K(\tau,x_0)$, it suffices to show that any neighbourhood of $0$ contains a point in $K(\tau,x_0)$ and a point outside. Let's fix $\epsilon>0$ and take the open ball $B:=\{x\in\mathbb R^n \,:\, \|x\|<\epsilon\}$. Since $0\in K(\tau,x_0)$, there exists $\alpha:[0,\tau]\to[-1,1]^m$ so that $$ 0 = X(\tau)x_0 + X(\tau)\int_0^\tau X^{-1}(s)N\alpha(s)ds $$ For any $T<\tau$ take the state $$ x(T) = X(T)x_0 + X(T)\int_0^T X^{-1}(s)N\alpha(s)ds $$ Then $x(T)$ is in $K(\tau,x_0)$, since it is in $K(T,x_0)$.
Let's choose $T$ so that $x(T)\in B$. That is possible, since, once fixed $x_0$, $t\mapsto x(t)$ is differentiable and thus locally Lipschitz. Hence we can bound $\|x(\tau)-x(T)\|$ as $$\|x(\tau)-x(T)\|\le L\|\tau-T\|$$ with $L$ that depends on $\tau$. We can thus chose $T$ such that $\|x(\tau)-x(T)\|<\epsilon$. This proves that in the $\epsilon$-neighbourhood of $x(\tau)$ there exists a point in $K(\tau,x_0)$. Take now $T>\tau$ and $\alpha'(s)$ that is any properly defined non-zero function that agrees with $\alpha$ on $[0,\tau]$. then $x(T)\notin K(t,x_0)$, for all $t\le \tau$, otherwise $x(\tau)$ would be reachable in less that $\tau$ seconds, that violates the hypothesis. Using the same argument as before you can chose $T>\tau$ so that $x(T)$ is in the $\epsilon$-neighbourhood of $x(\tau)$, thus proving that in such neighbourhood there exists a point that does not belong to $K(\tau,x_0)$. At this point, the claim should follow from the arbitrariness of $\epsilon$.
Maybe this is not a rigorous proof, but I hope that it can be an idea.