Here is a bit of Mathematica code that rather supports Joseph's conclusion.
Tet[phi_] := {{Sin[phi], 0, Cos[phi]}, {-Sin[phi], 0, Cos[phi]}, {0, Sin[phi], -Cos[phi]}, {0, -Sin[phi], -Cos[phi]}};
Rect[a_, phi_] := Module[{x, y, u, w}, {x, y, u, w} = Tet[phi]; Polygon[{a x + (1 - a) u, a x + (1 - a) w, a y + (1 - a) w, a y + (1 - a) u}]];
v = Subsets[Range[4], {3}];
Manipulate[ Graphics3D[{Opacity[0.2], Sphere[{0, 0, 0}, 1], Opacity[0.3], GraphicsComplex[Tet[phi], Polygon[v]], Opacity[0.8], Rect[a, phi]}], {phi, 0, Pi/2}, {a, 0, 1}]
Using parameters like a=0.9, phi=1.4 one obtains an elongated rectangle inscribed a flat tetrahedron, close to an equatorial plane. The maximal inscribed ellipse in this rectangle hardly is contained in any triangle that fits in the unit ball.
Edit:
The following improved version fits the largest possible ellipse in the rectangle and finds the smallest enclosing triangle. This actually shows that in all above cases a triangle $T' \subset B$ can be found. This now rather strengthens the belief that the statement is true.
(Note: the triangle should turn red if a counterexample occurs.)
Code:
Tet[phi_] := {{Sin[phi], 0, Cos[phi]}, {-Sin[phi], 0, Cos[phi]}, {0, Sin[phi], -Cos[phi]}, {0, -Sin[phi], -Cos[phi]}}; Rect[a_, phi_] := Module[{x, y, u, w}, {x, y, u, w} = Tet[phi]; Polygon[{a x + (1 - a) u, a x + (1 - a) w, a y + (1 - a) w, a y + (1 - a) u}]]; v = Subsets[Range[4], {3}];
ParPl[a_, phi_] := ParametricPlot3D[{a Sin[phi] Cos[t], -(-1 + a) Sin[phi] Sin[ t], (-1 + 2 a) Cos[phi]}, {t, 0, 2 Pi}, Boxed -> False, Axes -> False];
Tangent[a_, phi_, t_, lam_] := {a Sin[phi] Cos[t], -(-1 + a) Sin[phi] Sin[ t], (-1 + 2 a) Cos[phi]} + lam {-a Sin[phi] Sin[t], (1 - a) Cos[t] Sin[phi], 0};
Manipulate[ p1 = Tangent[a, phi, tau, Cot[tau]]; p2 = Tangent[a, phi, tau, -Sec[tau] - Tan[tau]]; f[tau_] = p1.p1 // Simplify; g[tau_] = p2.p2 // Simplify; bt = tau /. FindRoot[f[tau] == g[tau], {tau, 0.1, 10^-10, Pi/2}]; p1 = Tangent[a, phi, bt, Cot[bt]]; p2 = Tangent[a, phi, bt, -Sec[bt] - Tan[bt]]; p3 = {-1, 1, 1}*p2; col = If[Max[p1.p1, p2.p2, p3.p3] <= 1, Green, Red]; Show[Graphics3D[{Opacity[0.2], Sphere[{0, 0, 0}, 1], Opacity[0.3], GraphicsComplex[Tet[phi], Polygon[v]], Opacity[0.8], Rect[a, phi], col, Polygon[{p1, p2, p3}]}], ParPl[a, phi]], {{phi, Pi/4}, 0, Pi/2}, {{a, 0.7}, 0, 1}]
