Full problem statement:
Show $f_n=\frac{x^n}{1+x^n}, \ (0 \leq x \leq 1)$ converges pointwise on $[0,1]$.
If $f(x) = \lim_{n \to \infty} f_n(x)$, is there $N \in$ $\mathbb{N}$ s.t. $|f_n(x) - f(x)| \lt \frac{1}{4}, \ (n \geq N)$ for all $x \in [0,1]$ simultaneously?
I was able to prove $f_n$ converges pointwise to a function $f$ where,
$f(x) = 0$ for $0 \leq x \lt 1$
and
$f(x)= 1/2$ for $x=1$.
Now there is a theorem that states that if $f_n(x)$ is continuous for each $n$, and it converges uniformly to $f$, then $f$ must be continuous.
In this case $f$ is not continuous since it's a piecewise function with values $0$ and $1/2$. So the contrapositive statement means that $f_n$ cannot converge to $f$ uniformly.
So for any given $\epsilon \gt 0$, there is no $N$ s.t. $|f_n(x) - f(x)| \lt \epsilon, \ (n \geq N)$ for all $x$ simultaneously.
BUT!
Case 1 $(x = 1)$
$|f_n(x) - f(x)| = |0-0| \lt \frac{1}{4}, \ \forall n \in \mathbb{N}$
Case 2 $(0 \leq x \lt 1)$
$f(x) = 0$
So $|f_n(x) - f(x)| = \left| \frac{x^n}{1+x^n} - 0 \right|$ $\leq \left| \frac{x^n}{1} - 0\right| = |x^n| \lt \frac{1}{4}$ for a sufficiently large $n$ since $|x| \lt 1$. So let this index be $N$.
Now putting this all together, if we let $n \geq N$, then both cases are satisfied. So there does exist $N$ that works for $\epsilon = \frac{1}{4}$ $\ \ \ \forall x$ simultaneously.
So for $\frac{1}{4}$, it works. But notice that this would have worked for any given $\epsilon$.
So clearly I made a mistake but I am not able to pinpoint where.
What is my mistake?