How to prove that for all $\epsilon>0$, there exist $x_{1},\dots,x_{k}\in\left[0,1\right]$ such that $$ \left[0,1\right]\subset\bigcup_{i=1}^{k}\left[x_{i}-\epsilon,x_{i}+\epsilon\right]. $$ I found on Wikipedia that this property is called total boundedness.
My try: If $\epsilon=\frac{1}{6}$ then $x_k$ are: $\frac{1}{6},\frac{3}{6},\frac{5}{6}$. If $\epsilon=\frac{1}{8}$ then $x_k$ are: $\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8}$.
Just do it.
Just let the intervals be $[0, 2\epsilon], [2\epsilon, 4\epsilon], [4\epsilon, 6\epsilon]$ etc.
So $x_1= \epsilon$; $x_2 = 3\epsilon; x_3 = 3\epsilon; x_i =(2i -1)\epsilon$.
how many of these are there? well, we end up with $ 1\in [(2n-1)\epsilon, (2n+1)\epsilon$ so that $2n \epsilon > 1$ and $n > \frac 1{2\epsilon}$.
So just let $[0,1]\subset \bigcup\limits_{i=1;x_i = (2i-1)\epsilon}^{\frac 1 {2\epsilon}}[x_i - \epsilon, x_i + \epsilon]= \bigcup\limits_{i=1}^{\frac 1 {2\epsilon}}[(2i-2) \epsilon, 2i\epsilon] $
......
Or if that's a little vague or complex:
Notice for every $\epsilon > 0$ there exist an $n\in \mathbb N$ so that $\epsilon > \frac 1n > 0$.
[Have you proven that yet? It should be a basic result you are comfortable with.]
Then do your try: $x_i = \frac 1n, \frac 3n, \frac 5n, ..... \frac {2i-1}n$.
.....
Also. There is nothing wrong with overlap.
$[0, 2\epsilon],[\epsilon, 3\epsilon], [2\epsilon, 4\epsilon]....$ has more over lap. But that's okay. If the intervals were open (which would have to be the case if this were actually a prove about compactness-- which despite that tag it is not) you would need to have overlap.... and the math is actually easier.
$[0,1] \subset \bigcup\limits_{i=1}^{\frac 1\epsilon}[(i-1)\epsilon, (i+1)\epsilon]$.