The answer is $K=c^{-c}(1-c)^{-(1-c)}$. We will first show that this $K$ satisfies the condition. Let $x_i = c^{-i}$. Then for any positive real $M$, there exists $n$ such that $x_n\geq M$. Because \[\sum_{i=0}^{\infty}\frac{\left(x_{i+1}-x_i\right)^c}{x_i^{c+1}}=\sum_{i=0}^{\infty}\frac{(1-c)^cc^{-c}x_i^c}{x_i^{c+1}}=(1-c)^cc^{-c}\sum_{i=0}^{\infty}c^i = c^{-c}(1-c)^{-(1-c)},\]this seqeuence satisfies the condition, which shows that this particular $K$ satisfies the condition. To show that there does not exist $K<c^{-c}(1-c)^{-(1-c)}$ meeting the condition, we just need to show that there exists a positive real $M$ such that there does not exist any strictly increasing sequence satisfying the condition. Let $M$ be a sufficiently large positive real such that $c^{-c}(1-c)^{-(1-c)}(1-1/M)>K$. Then we just need to show that for any strictly increasing sequence $x_0,\ldots,x_n$ with $x_0=1$ and $x_n\geq M$, we have \[\sum_{i=0}^{n-1}\frac{\left(x_{i+1}-x_i\right)^c}{x_i^{c+1}}\geq c^{-c}(1-c)^{-(1-c)}\left(1-\frac{1}{M}\right).\]By the power mean inequality, \[\frac{\left(x_{i+1}-x_i\right)^c}{x_i^{c+1}}=\frac{x_{i+1}-x_i}{x_i}\cdot \frac{1}{x_i^c(x_{i+1}-x_i)^{1-c}} = c^{-c}(1-c)^{-(1-c)}\cdot\frac{x_{i+1}-x_i}{x_i}\cdot \frac{1}{\left(\frac{x_i}{c}\right)^c\left(\frac{x_{i+1}-x_i}{1-c}\right)^{1-c}}\]\[\geq c^{-c}(1-c)^{-(1-c)}\cdot\frac{x_{i+1}-x_i}{x_i}\cdot \frac{1}{x_{i+1}}=c^{-c}(1-c)^{-(1-c)}\left(\frac{1}{x_i}-\frac{1}{x_{i+1}}\right). \]Hence by summing over $i$, \[\sum_{i=0}^{n-1}\frac{\left(x_{i+1}-x_i\right)^c}{x_i^{c+1}}\geq c^{-c}(1-c)^{-(1-c)}\left(\frac{1}{x_0}-\frac{1}{x_{n}}\right)\geq c^{-c}(1-c)^{-(1-c)}\left(1-\frac{1}{M}\right)>K.\]Therefore any $K$ that is less than $c^{-c}(1-c)^{-(1-c)}$ does not satisfy the condition.