#JavaScript (ES6), 71 bytes
Takes input as an array of numbers. Falsy values are 0 or NaN. Truthy values are strictly positive integers.
a=>a[1]*a.every(p=n=>(373>>n|373>>p)&n+p!=10|a[-p-(p=n)>>1]?a[-n]^=1:0)
###Test cases
let f =
a=>a[1]*a.every(p=n=>(373>>n|373>>p)&n+p!=10|a[-p-(p=n)>>1]?a[-n]^=1:0)
console.log('[Truthy]')
console.log(f([1,2]))
console.log(f([1,6]))
console.log(f([5,8,2]))
console.log(f([1,2,3,5,7]))
console.log(f([1,5,7,8,2]))
console.log(f([4,2,6,3,1]))
console.log(f([1,5,7,8,4,2]))
console.log('[Falsy]')
console.log(f([1]))
console.log(f([1,3]))
console.log(f([3,1]))
console.log(f([3,3]))
console.log(f([1,3,7]))
console.log(f([1,5,4,1]))
console.log(f([1,9,7,3,5]))#Original answer, 90 bytes
Saved 2 bytes by mixing different falsy/truthy values, as suggested by @HermanLauenstein
Takes input as an array of numbers. Falsy values are 0 or NaN. Truthy values are strictly positive integers.
a=>a[b=1]*a.every(n=>!(p=b,b=1<<n,j=-'085645205005'[(p^b)*60%198%37])|p<2|a[j]?a[-n]^=1:0)
###Test cases
let f =
a=>a[b=1]*a.every(n=>!(p=b,b=1<<n,j=-'085645205005'[(p^b)*60%198%37])|p<2|a[j]?a[-n]^=1:0)
console.log('[Truthy]')
console.log(f([1,2]))
console.log(f([1,6]))
console.log(f([5,8,2]))
console.log(f([1,2,3,5,7]))
console.log(f([1,5,7,8,2]))
console.log(f([4,2,6,3,1]))
console.log(f([1,5,7,8,4,2]))
console.log('[Falsy]')
console.log(f([1]))
console.log(f([1,3]))
console.log(f([3,1]))
console.log(f([3,3]))
console.log(f([1,3,7]))
console.log(f([1,5,4,1]))
console.log(f([1,9,7,3,5]))###Commented
a => // given the input array a[]
a[b = 1] * // test whether a[1] exists; initialize b to 1
a.every(n => // for each digit n in a[]:
!(p = b, // p = previous value of b
b = 1 << n, // b = bitmask with the n-th bit set (0-indexed)
j = -'085645205005'[ // find j = negated value of the 'in-between digit'
(p ^ b) // by XOR'ing b with p
* 60 % 198 % 37 // and applying the hash function (see below)
] // end of table lookup
) | // if j is zero
p < 2 | // or p is less than 2 (meaning this is the 1st iteration)
a[j] ? // or the (-j)-th digit was already visited:
a[-n] ^= 1 // mark the n-th digit as visited or fail if it already was
: // else:
0 // force every() to fail
) // end of every()
###Hash function
We're using the commutative function H() defined by:
H(a, b) = H(b, a) = ((1 << a) ^ (1 << b)) * 60 % 198 % 37
There are 8 combinations of keys that result in an 'in-between key' to test:
a | b | 1 << a | 1 << b | XOR | * 60 | MOD 198 | MOD 37
---+---+--------+--------+-----+-------+---------+--------
1 | 3 | 2 | 8 | 10 | 600 | 6 | 6 --> 2
1 | 7 | 2 | 128 | 130 | 7800 | 78 | 4 --> 4
1 | 9 | 2 | 512 | 514 | 30840 | 150 | 2 --> 5
2 | 8 | 4 | 256 | 260 | 15600 | 156 | 8 --> 5
3 | 7 | 8 | 128 | 136 | 8160 | 42 | 5 --> 5
3 | 9 | 8 | 512 | 520 | 31200 | 114 | 3 --> 6
4 | 6 | 16 | 64 | 80 | 4800 | 48 | 11 --> 5
7 | 9 | 128 | 512 | 640 | 38400 | 186 | 1 --> 8
For all other possible pairs (a, b), H(a, b) doesn't collide with any of the indices computed above.