On closed simple curve with curvature at most 1 I am looking for the reference to the following theorem.
I have to apply a similar statement, and it would be nice to trace the source.
Please note, I know few proofs in fact it is Problem 3 in my collection of exercises.

Theorem. Let $\gamma$ be a closed simple plane curve with curvature at most 1. Then the  disc bounded by $\gamma$ contains a circle of radius $1$.

Here is an illustration for those who are lazy to read.

 A: I would check work from the 1980s on the cut locus being a tree on surfaces.  Consider the inward normal exponential map of the simple curve, take its cut locus, the latter is a tree, take a leaf of the tree.  This point will be at least distance 1 away from the boundary if the (signed) curvature of the curve is less than 1.  This works for all curves (convex or not).
For the cut locus, see for example Itoh, Jin-ichi, Essential cut locus on a surface. Proceedings of the Fifth Pacific Rim Geometry Conference (Sendai, 2000), 53–59, Tohoku Math. Publ., 20, Tohoku Univ., Sendai, 2001. 
See also Zamfirescu, Tudor, On the critical points of a Riemannian surface. Adv. Geom. 6 (2006), no. 4, 493–500. 
A: Here is summary of letter from Yurii Nikonorov, it gives complete answer.
The statement is called Pestov--Ionin theorem:


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*Пестов Г., Ионин В. О наибольшем круге, вложенном в замкнутую кривую // Докл. АН СССР. – 1959. – 127, № 6. – С. 1170–1172.


An alternative proof via curve shortening flow was given by Pankrashkin:


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*Pankrashkin, Konstantin An inequality for the maximum curvature through a geometric flow. Arch. Math. (Basel) 105 (2015), no. 3, 297–300. 

A: This is not an answer, since I don't know a reference. But I now remember the proof when the curve is known to be convex. Here's a sketch:
The trick is to put the disk of radius 1 so that it osculates the curve at a point where the curvature is largest. Now parameterize the curve as a function of the angle $\theta$ that the outer unit normal makes with the positive $x$-axis with the origin placed at the center of the disk. Define the support function $h$ as a function of theta to be the dot product of the corresponding point on the curve with the outer unit normal. First, note that the disk is inside the curve if and only if $h \ge 1$ for all $\theta$. It can be shown that $h$ satisfies the ODE
$$
h'' + h = \rho
$$
where $\rho$ is the reciprocal of curvature. The result now follows by the Sturm comparison theorem. This proof works nicely in the sense that it can be generalized to higher dimensions using the second fundamental form.
This all uses standard stuff from convex geometry, so I think it's likely it was known to Blaschke if not even earlier than that. I imagine that the proof above could be adapted to nonconvex curves, but I haven't tried. If so, I would imagine that this was also known a long time ago.
Finally, the Harvard graduate student who figured out the proof somehow knew about the support function and that it satisfies the ODE above. The rest of us knew only the more usual differential geometric definitions of curvature, which were useless for this question.
