[Mihai-Dorian Vidrighin][1] has suggested the following idea. (He's asked me to post it because he can't post images.)

For simplicity, assume that the setup is "tight" in that $E$ meets every face of $T$ and the vertices of $T$ are on the boundary of $B$. The generalisation should be straightforward.

If any vertices of $T$ lie in the plane of $E$ then the cross-section is already a triangle, so assume otherwise. Pick an arbitrary vertex of $T$ and draw a cone staring from there with $E$ (shown in red) as a base. This cone meets the opposite face of $T$ in a new ellipse $E'$ (shown as a dotted black curve):

![Tetrahedron with ellipse and cone][2]

$E'$ lies inside a triangle (the face of $T$), which itself lies inside a circle (the cross-section of $B$). By our simplifying assumption, $E'$ touches the edges of the triangle and the triangle's vertices are on the circle. Therefore [Poncelet's Porism][4] applies. Hence we can find a different triangle around $E'$ with one edge in the plane of $E$.

Define a new tetrahedron (shown in green) using the chosen vertex of $T$ and the new triangle. By construction it still contains the cone identified before, and in particular it still contains $E$. But the cross-section in the plane of $E$ is now a triangle (shown in thick black).

![alt text][3]
 

  [1]: http://www3.imperial.ac.uk/controlledquantumdynamics/people/students/cohortthree/mihai-dorianvidrighin
  [2]: http://s9.postimage.org/wny1rlowv/Untitled_2.png
  [3]: http://s17.postimage.org/8m2y5xdwv/Untitled_3.png
  [4]: http://mathworld.wolfram.com/PonceletsPorism.html