In this answer I will argue that this variational problem is an instance of a recurring theme, and that an overarching insight can be applied, which then serves as the proof that is sought after.
The first to hit upon that overarching insight was Jakob Bernoulli, the older brother of Johann Bernoulli. (It was Johann Bernoulli who had issued the Brachistochrone challenge to the mathematicians of his time.)
Using a differential approach Jacob Bernoulli managed to find independently that the cycloid is the solution to the Brachistochrone problem.
Jakob Bernoulli opened his discussion of the Brachistochrone problem with a lemma illustrated with diagram 1
Diagram 1
Lemma. Let ACEDB be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along AGFDB in a shorter time than along ACEDB, which is contrary to our supposition.
Another way of phrasing Jakob's Lemma is as follows: choose any subsection of the total curve. That subsection is in and of itself an instance of the Brachistochrone problem; the object traverses it in the smallest time. That property is valid down to infinitisemally small subsections.
The diagram I included is copied from the diagram that accompanied Jakob Bernoulli's original publication in the Acta Eruditorum, May 1697, pp. 211-217
The full content of the issues of the Acta Erutidorum is available on archive.org as scans of the pages.
Source of diagram 1
Returning now to the specific case that is the subject of your question:
In the image you supplied in your question: we can think of the left hand image, the diagram labeled (a), as an instance of the unit of operation of calculus of variations.
The unit of operation of calculus of variations consists of a triplet of points. For each instance of the unit of operation the outer points are treated as fixed points, and the middle point is shifted along a single coordinate; that shift is the variation. (The unit of operation of differential calculus is a pair of points. In the limit of the points of that pair being infinitisemally close together the line through those two points approaches the tangent to the curve.)
We can think of the right hand diagram, the diagram labeled (b), as a concatenation of instances of the unit of operation (with the triplets overlapping).
(b) represents a concatenation of three instances of the unit of operation.
In (b): the global transit time is an extremum if and only if all at once for the three instances of the unit of operation the transit time is an extremum.
The qualificiation 'all at once' refers to a condition of being concurrently satisfied. The condition must be satisfied for all the concatenated units of operation concurrently.
That reasoning is valid for any number N of layers; it extends to an infinite number of layers.
In a proof with equations you would be forced anyway to allow the number N to go to infinity, since that is the general case anyway. A proof that is valid for the number N infinitely large automatically covers all finite values of N too.
On my own website I offer an exposition of Calculus of Variations where I use the case of a soap film stretching between coaxial rings as motivating problem.
(A link to my website is available on my stackexchange profile page.)
The exposition uses interactive diagrams. For the soap film case I offer a diagram with 4 sliders. Moving the sliders changes the shape of the contour. By iterating over the four sliders the visitor converges onto the minimal surface.
All applications of Calculus of Variation have that common theme: concatenation of instances of the unit of operation. In the limit of making the instances of the unit of operation infinitely small the problem can be stated as a differential calculus problem.
