Cauchy by Definition I'm a bit confused as to how to show a sequence is Cauchy by definition. For example let $a_n = 1 - \frac{2}{\sqrt{n}}$. 
$$\left|\left( 1 - \frac{2}{\sqrt{n}}\right) - \left( 1 - \frac{2}{\sqrt{m}}\right)\right| = \left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right|$$
And let $m \geq n$. Then this is where I got lost in class: $|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}| \leq |\frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}|$.
Can someone explain why the last step is justified? Is it because $m \geq n \implies \frac{2}{\sqrt{m}} \leq \frac{2}{\sqrt{n}}$ and $\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}} \leq 0 \leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}$? But the inequality may not hold when taking absolute values of both sides...

In the situation you have, with $m\geq n$, we have
$$\left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right| = \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{m}} \leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}} = \left|\frac{2}{\sqrt{n}} +\frac{2}{\sqrt{n}}\right|$$
because 
$$-\frac{2}{\sqrt{m}} \leq 0 \leq \frac{2}{\sqrt{n}},$$
so adding $\frac{2}{\sqrt{n}}$ to both sides leads to the last inequality.
Alternatively,
$$\begin{align*}
\left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right| &\leq \left|\frac{2}{\sqrt{m}} \right| + \left|\frac{2}{\sqrt{n}}\right| &\text{(triangle inequality)}\\
&=\frac{2}{\sqrt{m}} + \frac{2}{\sqrt{n}}\\
&\leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}\\
&= \left|\frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}\right|.
\end{align*}$$
You prove a sequence $\{a_n\}$ is Cauchy "by definition" by showing that for every $\epsilon\gt 0$ there exists an $N$ (which will depend on $\epsilon$) such that for all $n,m\geq N$ we have $|a_n-a_m|\lt \epsilon$. 
