Is it possible to inscribe a regular tetrahedron in every convex body? Is it possible to inscribe at least one regular tetrahedron in every convex body?
 A: The first theorem of Section 4 of this paper, mentioned by J.M. in the comments, gives an affirmative answer, citing V.V Makeev, Inscribed simplices of a convex body (in Russian), Ukr.
Geom. Sb. 35 (1992), 47-49 = J. Math. Sci. 72 (1994) (4), 3189-3190, MR
95d:52006:

Theorem. Let $K\subset\mathbb{R}^n$ be a convex body. Then $K$ admits an inscribed similar copy of any prescribed simplex.

A: Suppose this is in $\mathbb{R}^3$, and of course it is necessary for the convex body to affinely span the whole $3$-dimensional space (since a regular tetrahedon within the body would space that whole space).  (Is that part of the definition of "body"?)  Then there are four affinely independent points $a,b,c,d$ within the convex body.  Consider the point $o=(a+b+c+d)/4$.  Let $p,q,r,s$ be the vertices of a regular tetrahedron centered at $o$, and the consider the points $o+\varepsilon(p-o)$, $o+\varepsilon(q-o)$, $o+\varepsilon(r-o)$, $o+\varepsilon(s-o)$, where $\varepsilon>0$.  These are vertices of a regular tetrahedron.  For $\varepsilon$ small enough, the weights $w,x,y,z$ such that $o+\varepsilon(a-o)=wa+xb+yc+zd$ should be close to $1/4$, hence all positive, and similarly for $b, c, d$. So those vertices should be in the convex hull of $\{a,b,c,d\}$, hence within the original convex body.
However, I wonder if I'm missing something?
Later edit: My suspicion is confirmed below: I missed one of the definitions.
A: Here is a simple proof.
Fix a coordinate system. Suppose that the body sits between $z=z_0$ and $z=z_1$.
For each $z_0 < z < z_1$, cut the body with a plane parallel to $xy$-plane. Inscribe an equilateral triangle in this intersection (to show it is possible, repeat this proof for two dimensions).
Let $G_z$ be the circumcircle of this triangle and $l_z$ be the length of the edge of the triangle, lets call the triangle $A_zB_zC_z$. 
Draw a semi-line parallel to the $z$ axis, going up, and starting at $G_z$. Let $L_z$ be the intersection of this semiline with the body (it's unique by convexity)$.
Let $f(z)=\frac{G_zL_z}{l_z}$. This ratio is continuous as a function in $z$.
It is easy to prove that 
$$\lim_{z \to z_0^+} f(z)= \infty \,,\, \lim_{z \to z_1^-} f(z)=0 \,.$$
Then, by the intermediate value theorem, there exists some $z_3$ so that $f(z_3)$ is exactly the ratio of the height of the regular tetrahedron/edge of the regular tetrahedron.
Now it is trivial to prove that $A_{z_3}B_{z_3}C_{z_3}L_{z_3}$ is a tetrahedron which satisfies the requirements.
Moreover, you can change the proof to show that you can inscribe such a tetrahedron so that its "basis" is parallel to any given plane.... 
