[Deleting previous answer, and starting over.]
This is a special case of a more general result (which I suspect must be known, but wasn't terribly difficult to derive independently) about when the endpoints of three concurrent segments lie on a conic (are "co-conicear"?).
Proposition. Suppose six distinct points, $P_0$, $P_1$, $P_2$, $Q_0$, $Q_1$, $Q_2$ are such that the lines $P_i Q_i$ are distinct and concur at a distinct seventh point, $R$. Write $p_i := \pm|RP_i|$, $q_i := \pm|RQ_i|$, and $\theta_i = \angle P_{i-1}RP_{i+1}$; for each $i$, we assign $p_i$ and $q_i$ opposite signs if $R$ is between $P_i$ and $Q_i$, and identical signs otherwise. (The $p_i$ and $q_i$ cannot be zero, nor equal.) Then, those six points lie on a common conic if and only if
$$ \sum_i \left(\frac{1}{p_i}+\frac{1}{q_i}\right)\sin\theta_i = 0$$
Proof. Take $R$ to be the origin, and assign coordinates to the points:
$$\begin{eqnarray*}
P_0 = p_0 (1, 0) \hspace{0.25in} P_1 = p_1(\cos\phi_1, \sin\phi_1) \hspace{0.25in} P_2 = p_2 (\cos\phi_2,\sin\phi_2)\\
Q_0 = q_0 (1, 0) \hspace{0.25in} Q_1 = q_1(\cos\phi_1, \sin\phi_1) \hspace{0.25in} Q_2 = q_2 (\cos\phi_2,\sin\phi_2)
\end{eqnarray*}$$
Substitute the coordinates into the conic template $A x^2 + 2 B xy + Cy^2 + 2 D x + 2 Ey + 1 = 0$ to get a system of six equations in five unknown coefficients. (Note that any such conic will not pass through the origin, or $R$ in the general context.) A computer algebra system helps to determine that the system imposes a dependency among the $p_i$, $q_i$, and $\theta_i$ ---for instance, evaluating the determinant in equation (8) of MathWorld's "Conic Section" entry, with the sixth point's coordinates entered into the top row--- that dependency can be expressed as the summation above, with (taking $\phi_0 := 0$) $\sin\theta_i := \sin(\phi_{i+1}-\phi_{i-1})$. Note that our assumptions provide that $p_i \ne 0$, $q_i \ne 0$, $p_i \ne q_i$, and that $\sin\theta_i\ne 0$.
To the problem at hand ...
The Proposition is clearly relevant, since pairs of a triangle's "sub-incenters" determine three segments that concur at the triangle's incenter. We just have to figure out the lengths and angles involved to run the "co-conicearity test".
First, some notation: Let $\triangle ABC$ have incenter $I$, and let $N_A$, $N_B$, $N_C$ be the points where angle bisectors $AI$, $BI$, and $CI$ meet sides $BC$, $CA$, and $AB$, respectively. (If you consult the first figure in MathWorld's "Incircle" entry, my $N$s are close to their $M$s ... which is why I chose "$N$" to denote them.) Let $P_{AB}$ be the incenter of $\triangle IAN_C$; the "$AB$" indicates that the triangle lies along $A$'s end of side $AB$. Likewise, let $P_{BC}$ and $P_{CA}$ be the incenters of $\triangle IBN_A$ and $\triangle ICN_B$; and let $Q_{BA}$, $Q_{CB}$, $Q_{AC}$ be the incenters of $\triangle IBN_C$, $\triangle ICN_A$, and $\triangle IAN_B$. (The concurrent segments are then $P_{CA}Q_{BA}$, $P_{AB}Q_{CB}$, $P_{BC}Q_{AC}$.)
Because $AI$, $BI$, and $CI$ bisect vertex angles, and each $P_\star I$ and $Q_\star I$ bisects an angle at $I$, finding (the sines of) the angles between segments is straightforward, as shown below.
Note. To save space, I write "$\theta_n$" for "$\frac{\theta}{n}$". Observe that $\alpha_n+\beta_n+\gamma_n = \pi_n$.
$$\begin{eqnarray*}
\sin\angle P_{BC}IP_{CA} &=& \sin(\angle BIN_B - \angle BIP_{BC} - \angle N_B IP_{CA} \\
&=& \sin\left( \pi - \frac{1}{2} \angle BIN_A - \frac{1}{2}\angle CIN_B \right) \\
&=& \sin\left(\frac{1}{2}(\alpha_2+\beta_2)+\frac{1}{2}(\gamma_2+\beta_2)\right) = \sin\left(\alpha_4 + 2\beta_4+\gamma_4\right) \\
&=& \sin\left( \pi_4 + \beta_4 \right)
\end{eqnarray*}$$
For the length information, we need to formulate the distance from a triangle's incenter to one of its vertices.
Let $\triangle ABC$ have edge lengths $a$, $b$, and $c$ (in the standard arrangement), and let $d$ be the triangle's circumdiameter (which of course, satisfies $d = \frac{a}{\sin\alpha} = \frac{b}{\sin\beta}=\frac{c}{\sin\gamma}$). Invoke the Law of Sines in $\triangle IBA$, and manipulate appropriately, to get the following
$$|IA| = \frac{|AB|\sin\angle IBA}{\sin(\pi-\angle IBA-\angle IAB)} = \frac{c\sin\beta_2}{\sin(\alpha_2+\beta_2)} = \frac{d\sin\gamma\sin\beta_2}{\cos\gamma_2}= 2 d \sin\beta_2\sin\gamma_2$$
(The use of the circumdiameter makes the symmetry clear.)
We want to apply that formula to, say, $\triangle IAN_C$, which has incenter $P_{AB}$. Within this triangle, the angles at $A$ and $N_C$ are, respectively $\alpha_2$ and (from looking at $ACN_C$) $\pi-\alpha-\gamma_2=\pi_2+\beta_2-\alpha_2$. The circumdiameter is
$$d_{AB} = \frac{|IA|}{\sin\angle N_C} = \frac{2d\sin\beta_2\sin\gamma_2}{\sin(\pi_2+\beta_2-\alpha_2)}=\frac{2d\sin\beta_2\sin\gamma_2}{\cos(\beta_2-\alpha_2)}$$
and we can compute
$$|IP_{AB}| = 2 d_{AB} \sin\frac{\alpha_2}{2}\sin\frac{\pi_2+\beta_2-\alpha_2}{2} = \frac{4d\sin\alpha_4 \sin\beta_2\sin\gamma_2\sin(\pi_4+\beta_4-\alpha_4)}{\cos(\beta_2-\alpha_2)}$$
That's a little ugly. If we conveniently scale $\triangle ABC$ so that $d\sin\alpha_2\sin\beta_2\sin\gamma_2=1$, then we have
$$\begin{eqnarray*}
|IP_{AB}| &=& \frac{2\sin(\pi_4+\beta_4-\alpha_4)}{\cos\alpha_4\cos(\beta_2-\alpha_2)} = \frac{\sqrt{2}\left( \cos(\beta_4-\alpha_4)+\sin(\beta_4-\alpha_4)\right)}{\cos\alpha_4 \left( \cos^2(\beta_4-\alpha_4)-\sin^2(\beta_4-\alpha_4)\right)} \\
&=& \frac{\sqrt{2}}{\cos\alpha_4 \left(\cos(\beta_4-\alpha_4)-\sin(\beta_4-\alpha_4)\right)} = \frac{1}{\cos\alpha_4\cos\left(\pi_4+\beta_4-\alpha_4\right)}
\end{eqnarray*}$$
Recall that the co-conicearity test requires reciprocating these lengths; convenient, indeed!
All we need to do now is compute the terms from the test's sum. With $I$ indisputably separating members of each $PQ$ pair, we know to assign opposite signs to the corresponding segment lengths, giving this factor
$$\begin{eqnarray*}
\frac{1}{|IP_{AB}|}-\frac{1}{|IQ_{CB}|} &=& \cos\alpha_4 \cos(\pi_4+\beta_4-\alpha_4)-\cos\gamma_4\cos(\pi_4+\beta_4-\gamma_4) \\
&=& \frac{1}{2}\left( \cos(\pi_4-2\alpha_4+\beta_4)-\cos(\pi_4+\beta_4-2\gamma_4)\right) \\
&=& -\sin(\pi_4-\alpha_4+\beta_4-\gamma_4)\sin(\gamma_4-\alpha_4)\\
&=& -\sin\beta_2 \sin(\gamma_4-\alpha_4)
\end{eqnarray*}$$
so that
$$\begin{eqnarray*}
\left( \frac{1}{|IP_{AB}|}-\frac{1}{|IQ_{CB}|} \right) \sin\angle P_{BC}IP_{CA} &=& -\sin\beta_2 \sin(\gamma_4-\alpha_4)\sin(\pi_4+\beta_4) \\
&=& \frac{1}{2} \sin\beta_2 \left( \sin\gamma_2 - \sin\alpha_2 \right)
\end{eqnarray*}$$
(The utter simplicity here makes me think that my approach has ignored a Putnamian insight.)
Clearly, the Proposition's cyclic sum vanishes, so that the six sub-incenters lie on a common conic.
Verification that the conic is, specifically, an ellipse is --for now-- left to the reader.
Best Answer
Re your PS: Pick two points $P,Q$ in an equilateral triangle $A,B,C$. project $A,B,C$ to the line $PQ$, giving $A',B',C'$. Say $A'$ is the "leftmost" and $B'$ the "rightmost" of the projections. Then $|PQ|\le |A'B'|\le |AB|$. (We see that by the same argument, two points in a polygon are at most as far apart as two vertices, i.e., as the maximal edge-or.diagonal.
Now to c: Let $ABC$ be our triangle. Pick suitable $r=\frac{\sqrt3-1}{2}\approx 0.366<\frac12$. Draw the circles with radius $r$ around $A,B,C$. In the "middle" there is room for a small equilateral triangle with vertices on these circles and side length $1-r\sqrt 3=r$. Verify that all seven parts in the figure have diameter $r$. Therefore, the pigeons strike back if we have eight points.
To visualize, in the following illustration all blue circles have radius $r$ and centre one of the nodes: