I have seen this proof for periodicity of Nim sequences of finite Subtraction games:
Consider the finite subtraction game subtraction$(a_1, a_2, \dots , a_k)$ and its nim-sequence. From any position there are at most k legal moves.Therefore the nimber $G(n) \leq k$ for all $n$. Define $a =\max\{a_i\}$. Since $G(n) \leq k$ for all $n$ there are only finitely many possible blocks of $a$ consecutive values that can arise in the nim-sequence. So we can find positive integers $q$ and $r$ with $a \leq q < r$ such that the $a$ values in the nim-sequence immediately preceding $q$ are the same as those immediately preceding $r$. Then $G(q) = G(r)$.
Could someone explain what the bold line means, how is it inferred?
The point is that $G(n)$ depends only on $G(n-1), G(n-2),\ldots, G(n-a)$, because all moves you can make from $n$ take you to one of these values. Therefore, if we have numbers $n>m$ such that $(G(n-1), G(n-2),\ldots, G(n-a))=(G(m-1), G(m-2),\ldots, G(m-a))$ (that is, $G(n-i)=G(m-i)$ for each $1\leq i\leq a$), then $G(n)=G(m)$, and now we have the same condition on $n+1$ and $m+1$, so $G(n+1)=G(m+1)$, and so on. This means if we can find such $n,m$ then $G(n+\ell)=G(m+\ell)$ for all $\ell\geq 0$, i.e. the sequence is eventually periodic with period dividing $n-m$.
Now we need to argue that we can always find such numbers $n,m$. For each $n\geq a$ in turn, look at the vector $(G(n-1), G(n-2),\ldots, G(n-a))$. There are at most $(k+1)^a$ possible vectors, since we know that each entry is between $0$ and $k$ inclusive. So if you try more than $(k+1)^a$ different values of $n$, you must eventually find an $n$ that has the same vector as some smaller number $m$.