Is there any hope to save this proof? I am trying to prove the following:

Let $x, y\ge 0$ be reals and $n\ge 1$ be a natural number such that $x^n<y$. Then there exists an $\varepsilon>0$ such that $(x+\varepsilon)^n<y$.

My attempt is this:
Let $x, y\ge0$ be reals and $n\ge1$.
Let $\varepsilon>0$. Then there exists a real $K>0$ such that
$$(x+\varepsilon)^n < x^n + K\varepsilon$$
as is easily shown by induction. Now,
$$x^n + K\varepsilon < y\impliedby \varepsilon < \frac{y-x^n}{K}$$
and since $x^n<y$, there does exist an $\tilde\varepsilon$ such that $0<\tilde\varepsilon<\frac{y-x^n}{K}$.
Issue with this proof: It is not ensured that the same $K$ will also work with such an $\tilde\varepsilon$, hence rendering the proof incomplete.

Question: Is there any way to "fix" this proof, preserving the original ideas? If not, can you suggest an alternate approach?
 A: Your proof is actually fine. For a fixed $x,y,$ and $n$ (note here that the question is asking about fixed values and not, for example, all $n=1,2,...$) simply choose
$$\epsilon=\frac{y-x^n}{2K}$$
where $K$ is determined by some function of the $x,y,$ and $n$. If you want it explicitly written out, we have
$$(x+\epsilon)^n=\sum_{j=0}^n \binom{n}{j}\epsilon^{j}x^{n-j}=x^n+\sum_{j=1}^n \binom{n}{j}\epsilon^{j}x^{n-j}=x^n+\epsilon\sum_{j=1}^n \binom{n}{j}\epsilon^{j-1}x^{n-j}$$
Now, to bound this sum we have
$$\sum_{j=1}^n \binom{n}{j}\epsilon^{j-1}x^{n-j}<\left(\left(1+\frac{1}{x}\right)^n-1\right)x^n$$
which is true for $0<\epsilon<1$ (this is analogous to your $K$ term but here it depends on $x,y,$ and $n$).

EDIT: To really drive home how $K$ is only a function of $x,y,$ and $n$. I will restate what we have shown while explicitly connecting it to $K$.
For $0<\epsilon<1$ we have
$$(x+\epsilon)^n<x^n+K\epsilon$$
where $K$ is given by
$$K=\left(\left(1+\frac{1}{x}\right)^n-1\right)x^n$$
(in fact, this $K$ does not even depend on $y$). Again, note that this does not depend on $\epsilon$.

Using this, simply choose
$$\epsilon=\frac{y-x^n}{\left(\left(1+\frac{1}{x}\right)^n-1\right)x^n}$$
Putting it all together
$$(x+\epsilon)^n=x^n+\epsilon\sum_{j=1}^n \binom{n}{j}\epsilon^{j-1}x^{n-j}<x^n+\epsilon\left(\left(1+\frac{1}{x}\right)^n-1\right)x^n$$
$$=x^n+\frac{y-x^n}{\left(\left(1+\frac{1}{x}\right)^n-1\right)x^n}\left(\left(1+\frac{1}{x}\right)^n-1\right)x^n=x^n+y-x^n=y$$
A: Following your idea, one can handle the case $x=0$ and $x>0$ separately.
Case 1. If $x=0$ and $0<y,$ take $\epsilon=\min(1/2,y)>0.$ Then one has $$\epsilon^n<y.$$
Case 2. If $x>0$ and $x^n<y$, take $$\epsilon=\min\left(\frac x 2,\frac{y-x^n}K\right)>0,\qquad (1)$$ where $$K=(2^n-1)x^{n-1}>0.$$ One proceeds to show that $$(x+\epsilon)^n<y.$$ By (1), $\epsilon<x$, hence one has by the binomial coefficients that $$(x+\epsilon)^n=x^n+\epsilon(nx^{n-1}+\cdots+\epsilon^{n-1})$$
$$<x^n+\epsilon(2^n-1)x^{n-1}=x^n+K\epsilon$$
$$\leq x^n+K\cdot \frac{y-x^n}K=y,$$ where the last inequality follows from (1). QED
A: I'm not sure what the original idea is that you'd like to preserve, but here's a very simple approach:
Since $x^n<y$ and taking $n$-th roots is monotonic on non-negative reals, we also have $x < y^{1/n}$. Hence, we can define
$$
\Delta := y^{1/n} - x > 0.
$$
You want to show there is an $\varepsilon > 0$ such that $(x+\varepsilon)^n < y$, which by the same reasoning is equivalent to $x+\varepsilon < y^{1/n}$. Using $\Delta$ this condition translates to $\varepsilon < \Delta$. Hence, any $0<\varepsilon<\Delta$ will do, for example $\varepsilon=\Delta/2$.
