Proof that if a Dirichlet character has two induced moduli then their gcd must also be an induced modulus

Let \chi be a Dirichlet character mod k and let d be any positive divisor of k. The number d is called an induced modulus of \chi if we have

\chi(a) = 1 whenever gcd(a, k) = 1 and a \equiv 1 (mod d)

I wanted to construct an intuitively straightforward proof for myself that if \chi is a character mod k, and if k_1 and k_2 are induced moduli for \chi, then d = gcd(k_1, k_2) must also be an induced modulus for \chi, and \chi(a) = 1. I managed to come up with a relatively simple proof as follows.

Suppose a \equiv 1 (mod d), with gcd(a, k) = 1. Since k_1 and k_2 are induced moduli for \chi, \chi must take equal values at numbers which are relatively prime to k and congruent modulo either k_1 or k_2. In other words we must have

\chi(a) = \chi (a + xk_1) = \chi(a + yk_2)

for any integers x, y such that

gcd(a + xk_1, k) = gcd(a + yk_2, k) = 1

This is only possible in general if \chi takes equal values at numbers which are relatively prime to k and congruent modulo some common divisor of k_1 and k_2 (otherwise in some cases it would be possible to find some x, y such that \chi (a + xk_1) \neq \chi(a + yk_2)). Since any common divisor of k_1 and k_2 must also divide their greatest common divisor d, it follows that \chi must take equal values at numbers which are relatively prime to k and congruent modulo d. Thus, since a \equiv 1 (mod d), we have

\chi(a) = \chi(1) = 1

and d = gcd(k_1, k_2) is an induced modulus for \chi if k_1 and k_2 are. QED