Proof of existence:

The equation $$$ax+mk=1$$$ always has solution if $$$\gcd(a,m)=1$$$ (Bézout's lemma), so $$$x\bmod m$$$ is a modular multiplicative inverse of $$$a$$$.

Proof of uniqueness:

Assume that $$$p,q\,(0\le p < q < m, q - p = k)$$$ are two modular multiplicative inverses of $$$a$$$. This means $$$ap\equiv aq\equiv 1\pmod{m}$$$. If $$$\gcd(a,m)=1$$$, then $$$p\equiv q\pmod{m}$$$. This implies $$$p=q$$$, which leads to a contradiction.

Consider $$$0 \cdot a, 1 \cdot a, 2 \cdot a, ..., (m - 1) \cdot a$$$ mod $$$m$$$. They are different, because if $$$a \cdot i = a \cdot j$$$, then $$$a \cdot (i - j) = 0$$$ mod $$$m$$$, which is impossible since $$$gcd(a, m) = 1$$$ and $$$|i - j| < m$$$. Then, there exists and only one $$$b$$$: such $$$a \cdot b = 1$$$ mod $$$m$$$.