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Let $\gamma_0$ be a smooth closed curve in $\mathbb{R}^2$, such that it has no self intersection. You can smoothly deform it in order to create a (still smooth) curve $\gamma_1$ that have some self intersections. It looks like that at some point during the deformation you get at least one self intersection such that the two parts of the curve intersecting are tangent, for example as a limiting case between having $0$ and $2$ self intersections (see the image below).

Examples of curves

However, despite the problem looking quite simple I was unable to find a proof of this claim, either in a book or by myself. Therefore I am searching a reference to a proof of this claim, if it exists (which seems likely to me as the problem looks kind of standard).

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I don't have a reference, though it seems a simple application of the transversality theorem. The correct framework for this problem is so-called regular homotopies. In simple terms, a regular homotopy is the correct notion of homotopy between immersions; in this case, the initial and final closed curve can be thought of as immersions of the circle $S^1$ to the plane $\mathbb{R}^2$, and then your "deformation", if you assume it was done smoothly (which is a minimal necessity condition I assume from now on), is a regular homotopy between these two immersions.

Let's say this regular homotopy be

$$ \gamma : S^1\times I \to \mathbb{R}^2\\ (s,t)\mapsto \gamma_t(s) $$

where, as usual, $I=[0,1]$.

Here $\gamma_0:S^1\to \mathbb{R}^2$ is (a choice of parametrization of) the initial closed loop, and $\gamma_1:S^2\to\mathbb{R}^2$ that of the final closed loop. Then the regular homotopy condition just says

$$ \frac{\partial \gamma_t(s)}{\partial s}\neq 0, \qquad\forall (s,t)\in S^1\times I. $$

Now think of the map

$$ f:(S^1\times S^1\setminus \Delta) \times I \to \mathbb{R}^2\times \mathbb{R}^2\times I\\ (s_1,s_2,t)\mapsto (\gamma_t(s_1),\gamma_t(s_2),t) $$

where $\Delta$ is the diagonal of $S^1\times S^1$. This is clearly smooth, as a composition of (the obvious) smooth maps. Now, the set of self-intersections during the deformation is given by $f^{-1}(\Delta\times I)$ (Here $\Delta$ is the diagonal of $\mathbb{R}^2\times \mathbb{R}^2)$. (My notation might be slightly misleading as this set consists of points of the form $(s_1,s_2,t)$, not the points $(s_1,s_2)$, where $\gamma_t(s_1)=\gamma_t(s_2)$, but it doesn't disturb this proof).

Then, if there is no tangential intersection during the deformation, then $f$ is transverse to $\Delta\times I$, therefore we can use the transversality theorem. Thus, by a simple dimension count, $f^{-1}(\Delta\times I)$ is a 1-dimensional submanifold of $(S^1\times S^1\setminus \Delta)\times I$, meaning it is a union of closed intervals and circles in the domain $(S^1\times S^1\setminus \Delta) \times I$. But we assumed the initial immersion has no self-intersection, meaning there is no point $(s_1,s_2,0)$ in $f^{-1}(\Delta\times I)$. Hence, if you consider the connected component in $f^{-1}(\Delta\times I)$ containing the two self-intersection points at time $t=1$(these points exist by the OP's assumption that the end curve has self-intersections), then it must have a point tangential to $t=\mathrm{constant}$, which is impossible. Hence, there must be a tangential self-intersection during the deformation.

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