Understanding the proof that specific sequence does not have limit In mathematical analysis book, there is example of sequence $a_n$: $1$, $0$, $\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{2}{3}$, $\frac{1}{4}$, $\frac{3}{4}$, $...$
Odd numbers are part of sequence: $a_n = \frac{1}{n}$, $\lim\limits_{n\to \infty} a_n = 0$
Even numbers are part of sequence: $a_n = 1 - \frac{1}{n}$, $\lim\limits_{n\to \infty} a_n = 1$
Assuming since two subsequences have two different limits, whole sequence cannot have limit.
This one is clear to me, but where I am lost is explanation or an attempt to find that $\epsilon_0\in\mathbb{R}$ exists:
$$|a_\tilde{n} - a| \geq \epsilon_0$$
i.e. there exists least one $\epsilon_0\in\mathbb{R}$ disproving definition of limit:
$$|a_n - a| < \epsilon$$
for $a,\epsilon\in\mathbb{R}$, $a$ is limit of sequence $a_n$, $\forall n > n_0(\epsilon)$; $n,n_0\in\mathbb{N}$.
To show that sequence does not have limit, there is first proposition that for $\epsilon_0$, where for each $n\in\mathbb{N}$ there exist $\tilde{n} > n, \tilde{n}\in\mathbb{N}$ such as $|a_\tilde{n} - a| \geq \epsilon_0$, we choose number for $\epsilon_0$:
$$\epsilon_0 = \frac{1}{2}\max(|a|, |a - 1|)$$
For 1st assumption $|a - 1| \geq |a|$, because:
$$a_{2k} = 1 - \frac{1}{k} \quad \textrm{and} \quad \frac{1}{k} \to 0$$
there exists $k_0(\epsilon_0)\in\mathbb{N}$, where for $k > k_0(\epsilon_0)$ is always:
$$|a_{2k} - 1| < \epsilon_0, k\in\mathbb{N}$$
and therefore:
$$|a_{2k} - a| = |a_{2k} - 1 + (1 - a)| \geq |1 - a| - |a_{2k} - 1| \geq 2\epsilon_0 - \epsilon_0 = \epsilon_0$$
(let's call this inequality $A$)
Therefore for any $n$ therefore exists $ñ$ as even number, which has to satisfy following conditions:
$$ñ > n \quad \textrm{and} \quad ñ > 2k_0(\epsilon_0)$$
For example:
$$ñ = 2\max(n, k_0(\epsilon_0))$$
Then:
$$a_{2k}\in \langle1 - \epsilon_0, 1\rangle, k > k_0(\epsilon_0)$$
and therefore difference between $a_\tilde{n}$ and $a$ has to be at least $\epsilon_0$.
For 2nd assumption $|a - 1| \leq |a|$ would turn out by analogy, that $a$ has to be different to at least $|a|/2$ members of $a_{2k - 1}$ for $k$ big enough.

My question is, or rather I am stuck on how we got to the inequality $A$ in the first place based on previous assumptions and from that rest of assumptions that seem to stem from it.

I am sharing photos as well as link to pdf source for this example (in Czech language):


https://www2.karlin.mff.cuni.cz/~halas/MA/MA1/kopacek_-_mat._analyza_nejen_pro_fyziky_1.pdf

PS: Let me know if something there is not clear, I will try to make it more understandable if needed.

Every $\epsilon_0<1/2$ will do. Indeed, if there was an $a$ and an $n_0$ such that $|a_n-a| \le \epsilon_0$ for all $n>n_0$, then for $k>n_0$ we have
$|1-1/k-a|<\epsilon_0$ and $|1/k-a|<\epsilon_0$. By the triangle inequality, this implies that $|1-2/k|<2\epsilon_0$ which fails to hold for large $k$ if $\epsilon_0<1/2$.
