A geometric/intuitive proof of why shortest distance between two non-intersecting curves lie along the normal I saw this post but it was too technical for me.
I am looking for proof similar to what is shown in this Quora post by Nicholas Halderman. The part I found confusing/ hard of the proof was where he proves the lemma:'a small displacement in this direction along this curve will result in a point closer to B.',  I am looking for alternate geometric proof/ a more elaborate explanation on what he has done in the post.
 A: A geometric argument, assuming that a shortest segment exists:
Take any point $A$ on the first curve $\gamma_1$. In order to find the point nearest to $A$ on the second curve $\gamma_2$ draw a small circle centered at $A$ and blow it up until it meets, in fact touches, the second curve $\gamma_2$ at some point $A'$. From all points on $\gamma_2$ this point $A'$ is nearest to $A$. The segment $A A'$ is  a radius of the circle, hence orthogonal to the common tangent of the circle and $\gamma_2$ at $A'$.

Of course this works also in the other direction. It follows that for any segment from $\gamma_1$ to $\gamma_2$ that is not orthogonal on both curves  the above circle construction  finds a shorter one.
A: If curves are continuous and differentiable, take parametrizations $c_1(t)$ and $c_2(s)$ of the curves. We know that at the minimum points of the squared distance $\|c_1(t) - c_2(s)\|^2$ the derivative in each variable will be $0$. We get
$$ \frac{d}{dt}\|c_1(t) - c_2(s)\|^2 = 2(c_1(t) - c_2(s))\cdot c_1'(t) = 0 $$
That is, the points minimizing the distance satisfy $c_1(t) - c_2(s)$ being normal to $c_1'(t)$, which is parallel to the curve $c_1$ at $c_1(t)$. The same argument deriving the formula above for $s$ will show that $c_1(t) - c_2(s)$ is also normal to $c_2$ at $c_2(s)$.
A: Take curves $\gamma_1,\gamma_2$ and let the points at minimum distance be $P_1,P_2$.
Suppose the tangents to $\gamma_1$ at $P_1$ and to $\gamma_2$ at $P_2$ are not parallel.  Then they meet somewhere; let's say they meet further to the "left" along each tangent. That means we can draw parallel lines $\ell_1$ through $P_1$ and $\ell_2$ through $P_2$, such that as you go to the "left", the tangents are closer than $\ell_1$ and $\ell_2$. But points on $\gamma_1$ sufficiently close to $P_1$ are the same side of $\ell_1$ as the tangent is, and similarly for $\gamma_2$, meaning that going a small distance along each curve to the "left" gives two points which are closer together.
Now suppose the tangents are parallel but the normals do not coincide. Say the line segment $P_1P_2$ is to the "left" of the normal to $\gamma_1$ at $P_1$, and so to the "right" of the normal to $\gamma_2$ at $P_2$. Now we can draw parallel lines $\ell'_1,\ell'_2$ through $P_1$ and $P_2$ which are perpendicular to $P_1P_2$, and if we move a small amount to the "left" of $P_1$ along $\ell'_1$ and to the "right" of $P_2$ along $\ell'_2$, we get closer together. But as we move "left" from $P_1$ the tangent lies on the side of $\ell'_1$ which is closer to $\ell'_2$ and similarly for moving "right" from $P_2$. Since the curves lie between the tangents and $\ell'_1,\ell'2$ if we move a sufficiently small distance, this means they are even closer together.
