I think the figure should be a web. The problem was taken from a Turkish book, but for the above figure the condition in the problem is not met: If the vertices of the $k$-th pentagon are arranged in a clockwise manner as $5(k-1)+1, 5(k-1)+2, 5(k-1)+3, 5(k-1)+5, 5(k-1)+4$, the maximum difference between neighboring points will be at most $3$." The statement in the question is incorrect. The graph in the question is most likely as below: [asy][asy] real u=0.25cm; pair fP1, fP2, fP3, fP4, fP5; pair lP1, lP2, lP3, lP4, lP5; for(int i = 0; i < 4; ++i) { real v = u*(i+1); pair P1 = dir(90+0*72 - 18)*(0,v); pair P2 = dir(90+1*72 - 18)*(0,v); pair P3 = dir(90+2*72 - 18)*(0,v); pair P4 = dir(90+3*72 - 18)*(0,v); pair P5 = dir(90+4*72 - 18)*(0,v); dot(P1);dot(P2); dot(P3);dot(P4);dot(P5); path p = P1--P2--P3--P4--P5--cycle; draw(p); if(i == 0){ fP1 = P1; fP2 = P2; fP3 = P3; fP4= P4; fP5 = P5; } if(i == 3) { lP1 = P1; lP2 = P2; lP3 = P3; lP4= P4; lP5 = P5; } } draw(fP1--lP1);draw(fP2--lP2);draw(fP3--lP3);draw(fP4--lP4);draw(fP5--lP5); [/asy][/asy] $1$ must be one of the vertices of the innermost or outermost pentagon. Otherwise, $1$ would have four neighbors. Since one of them should be at least $5$, the difference $5-1=4$ would be greater than $3$, making the statement in the question inevitable. The innermost and outermost pentagons are actually symmetric. (You can think of the planar pentagons in space as four identical pentagons on four parallel planes.) Let's assume that $1$ is written on one of vertices of the outermost pentagons. $2, 3, 4$ will be written on neighboring vertices. Since $2, 3, 4$ are not adjacent to each other, $2$ must have at least two neighbors with unwritten numbers. One of them must be at least $6$. In this case, $6-2=4$ would be greater than $3$, so the existence of at least two neighboring vertices with differences greater than $3$ is inevitable.