Does a small-area sphere in a 3-manifold bound a small ball? Let $M$ be a closed riemannian 3-manifold. I think that the following fact should be true and should have a relatively simple proof, but I cannot figure it out.

For every $\varepsilon>0$ there is a $\delta>0$ such that every smooth 2-sphere in $M$ of area smaller than $\delta$ bounds a ball of volume smaller than $\varepsilon$.

Roughly, small-area spheres must bound small-volume balls. 
Note that: 


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If $M\neq S^3$ then $M$ contains spheres that bound regions of arbitrarily small volume that are not balls (just take a spine of $M$ and small regular neighborhoods of it).  

* It suffices to prove that the 2-sphere is contained in a small-volume ball and invoke Alexander theorem.

* The same fact stated for 3-spheres in $S^4$ would imply the (open) Schoenflies problem (every 3-sphere bounds a 4-ball), since every 3-sphere in $S^4$ can be shrinked to have arbitrarily small area.

* It is not true in general that a torus of small area is contained in a ball (pick neighborhoods of a homotopically non-trivial knot).

 A: This is a direct corollary of Federer -Flemming deformation Lemma saying that vary small area sphere can be homotoped to 1-skeleton of the fixed triangulation of the manifold. The dimension assumption is irrelevant. Proof of this Lemma can be found somewhere in Federer's book (I will chase down the precise reference when I can).
A: In their paper "The classical Plateau problem and the topology of 3-manifolds", Meeks and Yau claim that for any fixed closed Riemannian 3-manifold $M$ there is a lower bound for the area of non-trivial two-spheres.  Furthermore, the least area such is embedded (or double covers an essential $RP^2$.)
There are many ways to use this to answer the question.  Here is one possibility: 
We first induct on $s(M)$: the number of essential spheres in a maximal system in $M$.  If $s(M) = 0$, then $M$ has universal cover $R^3$ or $S^3$, and we are done, using the proof of Alexander's theorem, say.  (This step is not obvious, but let's move along.)
Suppose that $S$ is a minimal area essential sphere in $M$, and so is embedded.  Let $T$ be the given sphere with small area, which is necessarily inessential.  If $T \cap S$ is not generic, then move $T$ slightly to make it so.  We now induct on $|T \cap S|$.  If $T \cap S = \emptyset$ then we can cut along $S$ and cap off with a ball to get a manifold $M'$.  Note that $s(M') < s(M)$.  (We must also check that the area of a smallest essential sphere in $M'$ is greater than the area of $S$.)
If $|T \cap S| \neq \emptyset$, then let $\alpha$ be an innermost curve of intersection.  Thus $\alpha$ bounds a disk in $S$ that has area less than either disk it bounds in $T$.  (This uses the fact that a surgery of an essential sphere yields at least one essential sphere, and the minimality of $S$.)  So we may surger $T$ to get a pair of inessential spheres.  (This is because both of them have area less than that of $T$.)  This reduces $|T \cap S|$ and we are done.
