Uniform convergence of $f_n(x)=\left(x-\frac{1}{n}\right)^2$ for $x \in [0,1]$ Could someone please verify this?

Let $f_n(x)=\left(x-\frac{1}{n}\right)^2$ for $x \in [0,1]$.
(a) Does the sequence $(f_n)$ converge pointwise on the set $[0,1]$? If so, give the limit function.
(b) Does $(f_n)$ converge uniformly on $[0,1]$? Prove your assertion.

Let $x \in [0,1]$. Then,
\begin{equation}
f_n(x) = \left(x-\frac{1}{n}\right)^2
\end{equation}
Using the fact that $\lim \frac{1}{n} = 0$, along with limit theorems, we get
\begin{equation}
\lim f_n(x) = x^2
\end{equation}
Therefore, $f_n$ converges pointwise to $f(x) = x^2$ for all $x \in [0,1]$.
$f_n$ also converges uniformly to $f$ in $[0,1]$. To prove this assertion, let $\epsilon > 0$. Also, let $n > \frac{2}{\epsilon}$. Then,
\begin{eqnarray}
&n& > \frac{2}{\epsilon} \\
&\Rightarrow& n > \frac{2x}{\epsilon} \\
&\Rightarrow& 2x < n \epsilon \\
&\Rightarrow& \frac{2x}{n} < \epsilon \\
&\Rightarrow& \frac{2xn}{n^2} < \epsilon \\
&\Rightarrow& \left|\frac{1-2nx}{n^2}\right| < \epsilon
\end{eqnarray}
Note that the last implication holds true, since $$\frac{2xn}{n^2} > 0$$
This implies that $$\forall n > \frac{2}{\epsilon}, |f_n(x) - f(x)| < \epsilon$$
This completes the proof.

I don't think that you are justifying the last implication properly. Note you don't know the sign of $1-2nx$. What you can do is choose $n$ so that $1/n^2<\epsilon/2$, $2x/n<\epsilon/2$, and then use the triangle inequality. 
