Mihai-Dorian Vidrighin 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 http://s9.postimage.org/wny1rlowv/Untitled_2.png
$E'$ 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 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).
