I am uncertain whether my argument to disprove the uniform convergence of a sequence of holomorphic functions are correct.
Show that the sequence of holomorphic functions $f_n$ given by $f_n(z)=z^{2n}-z^{n+1}$ converges uniformly on every compact subset $K$ of the open unit disk $D$, but does not converge uniformly on $D$.
Using the representation $z=r e^{i \varphi}, \ r \in ]0,1[, \varphi in [0,2 \pi[ $ one sees that the sequence converges pointwise to $0$ since $|r|<1$.
In order to disprove the uniform convergence on $D$ I need to show
$$ \exists \varepsilon >0 \ \forall N \in \mathbb{N} \ \exists n \geq N \ \exists z \in D: |z^{2n}-z^{n+1}|< \varepsilon $$
Looking at the sequence on the interval $]-1,1[$ it seems intuitively clear that this $(f_n)$ cannot be uniformly convergent. I tried to solve this by considering $(f_n)$ on the interval ]0,1[ and find a contradiction.
Suppose that $(f_n)$ does converge uniformly on $D$. Then it converges uniformly on $]0,1[$. Let $1>\varepsilon >0$ for a suitable $\varepsilon$. Then there is $N \in \mathbb{N}$ such that for all $n \geq N$ and all $z \in D$ it follows $|z^{2n}-z^{n+1}| < \varepsilon$. Since the sequence $(a_n)_{n \in \mathbb{N}}, a_n:=|r^n-r|$ does not converge to $0$ for $r \in ]0,1[$ we may choose $n_0 \geq N$ and $r \in ]0,1[$ such that
\begin{align*} 1-\frac{1}{2n_0} & \leq r \\ \frac{|r^{n_0}-r|}{2} &> \varepsilon. \end{align*}
Since $r \in ]0,1[$ it follows from Lemma 1 (see below) that
\begin{align*} \frac{1}{2} &\leq r^{n_0} \tag{1} \\ \frac{|r^{n_0}-r|}{2} &> \varepsilon \tag{2}. \end{align*}
So
\begin{align*} \varepsilon &> |r^{2 n_0}-r^{n_0+1}| \\ &= r^{n_0} |r^{n_0}-r| \\ &\underset{\substack{(1)}}{\geq} \frac{|r^{n_0}-r| }{2} \\ &\underset{\substack{(2)}}{=} \varepsilon. \end{align*}
Then $\varepsilon> \varepsilon$ which is a contradiction. Hence $(f_n)$ does not converge uniformly on $D$.
I am however in doubt whether I can actually choose a specific $\varepsilon$ such that the inequality (2) holds. So I am trying to find an alternative proof. But I do not see how to choose a suitable $\varepsilon$ since the term $|r^{n_0}-r| $ is giving me trouble.
$\underline{\text{Lemma 1:}}$ Let $r \in ]0,1[$ and $n_0 \in \mathbb{N}$ such that $r \geq 1-\frac{1}{2 n_0}$. Then $\frac{1}{2} \leq r^{n_0}$.
$\underline{\text{Proof:}}$ Since $r \in ]0,1[$ it follows $1-r \geq -1$. Then Bernoulli's inequality yields
\begin{align*} r^{n_0}=(1-(1-r))^{n_0} \geq 1-n_0(1-r). \tag{*} \end{align*}
Since $r \geq 1-\frac{1}{2 n_0}$ we have
\begin{align*} r \geq 1-\frac{1}{2 n_0} & \Leftrightarrow \frac{1}{2n_0} \geq 1-r \\ & \Leftrightarrow n_0 (1-r) \leq \frac{1}{2}. \\ \end{align*}
Thus $\frac{1}{2}=1-\frac{1}{2} \leq 1-n_0 (1-r)$. This implies
\begin{align*} \frac{1}{2} \leq 1-n_0 (1-r) \underset{(*)}{\leq} r^{n_0}. \end{align*}
So $\frac{1}{2} \leq r^{n_0}$.
There is a much simpler way of doing this. If $K$ is a compact subset of $D$ then there exists $r \in (0,1)$ such that $|z| \leq r$ for all $z \in K$. Now $|f_n(z)| \leq r^{2n}+r^{n+1}$ for all $z \in K$ which proves uniform convergence on $K$. The convergence is not uniform on $D$ because $f_n(1-\frac 1n) \to \frac 1 {e^{2}} -\frac 1 e \neq 0$.