I am interesting in the proof of that every rational number cannot have in decimal form infinite number of digits that don't repeat (or the other way around).
So, then is enough to prove following statement: For any $n \in \mathbb{N}$ rational number $\frac{1}{n}$ can be represented in decimal form such that it's digits, if there are infinitely many, are repeating.
If this is true, then it is true for any $\frac{m}{n} = \frac{1}{n} + \frac{1}{n} + ... + \frac{1}{n}$ ($m$ times).
If you recall the long division algorithm, when you are in the fractional part, you consider the current remainder, append a zero and divide the new number, giving a new remainder.
Notice that the new remainder is only a function of the current remainder. As all remainders are smaller than the divisor, you will inevitably observe the same remainder after some time, and this starts an unlimited repetition.
E.g., for the fraction $\dfrac{722}{63}$, the remainders (past the integer part of the quotient) are
$$29,38,2,20,11,47,\color{green}{29}$$
The remainders follow the recurrence
$$r_{n+1}=(10\,r_n)\bmod d.$$
Simultaneously, the digits of the quotient are
$$\left\lfloor\frac{10\,r_n}d\right\rfloor,$$
$$4,6,0,3,1,7,\color{green}4.$$