$x_n = \dfrac {u_n + \cdots + u_{2n}} n$
$u_n \to 0$, so for every $\varepsilon > 0$ there is $N$ such that for every $n>N$ we have $|u_n| < \varepsilon$.
For every $\varepsilon > 0$, use (1) on $\varepsilon/2$ to get such an $N$.
Then whenever $n>N$, $x_n = \dfrac {u_n + \cdots + u_{2n}} n < \dfrac {\varepsilon/2 + \cdots + \varepsilon/2} n = \dfrac{n+1}{2n}\varepsilon < \varepsilon$. Similarly, $x_n > -\varepsilon$. Therefore, $|x_n| < \varepsilon$. $\square$

$x_n = \dfrac {u_n + \cdots + u_{2n}} n$
$u_n \to 0$, so for every $\varepsilon > 0$ there is $N$ such that for every $n>N$ we have $|u_n| < \varepsilon$.
For every $\varepsilon > 0$, use (1) on $\varepsilon/2$ to get such an $N$.
Then whenever $n>N$, $|x_n| = \left|\dfrac {u_n + \cdots + u_{2n}} n\right| < \dfrac {\varepsilon/2 + \cdots + \varepsilon/2} n = \dfrac{n+1}{2n}\varepsilon \le \varepsilon$. $\square$

Let $a, b, c, d \in \Bbb R$ be such that $ad − bc \neq 0$. Let $\mathcal R$ be the region in the plane bounded by
$|ax + by| \leq \alpha $ and $|cx + dy| \leq \beta$. Prove that the area of $\mathcal R $ is $\frac{4\alpha \beta }{|ad − bc|} $