Is it mathematical correct proof? Need rigorous proof of uniformly tight sequence. I need to prove that 

$P(\|X_n\| > M_n) \to 0$ as $n \to \infty$ for every $M_n \to \infty$

leads to 

$P(\|X_n\| > M) < \varepsilon $ for every $N \leq n$

I do the following:
I say that it is obvious that mostly for any $i$:
$M_i<M_{i+1}$, where $i=1,\ldots,\infty$
That is why the probabilities for different values of $M$ obey this inequality: 
$$P(\|X_n\| > M_i) > P(\|X_n\| > M_{i+1}) > \cdots > 0$$

Hence, we can state that there exist some $\varepsilon$ for which there
  exist $M$ such that using it, $\varepsilon$ can be chosen such that
  $P(\|X_n\| > M) < \varepsilon $

But I think this Hence, we can state... is not really mathematical proof. Which steps mathematically are omitted by me for this proof to be mathematically correct? Really do not want to skip any step and want to make this proof as meticulous as possible.
 A: Do you mean that "for all $\epsilon>0$" (instead of "there exists some $\epsilon>0$"), there exists some $M$ such that $\mathbb{P}(\Vert X_n\Vert >M)<\epsilon$ for large enough $n$? Because otherwise the proof is pretty simple, just pick $\epsilon=2$

A possible proof is by contradiction. Suppose that 

Hence, we can state that there exist some $\epsilon>0$ for which there exist $M$ such that there is some $N$ such that $n\geq N\Rightarrow \mathbb{P}(\Vert X_n \Vert >M)<ϵ$

is false. Then, there is a certain $\epsilon>0$ such that for all $M$, for all $n$, there is $k\geq n$ such that $\mathbb{P}(\Vert X_k \Vert >M)\geq ϵ$ 
Then, we construct recursively the following sequence $M_n$. $M_1=1$, and, for $M_n$, let
\begin{equation}
M_{n+1}=\begin{cases}
M_{n} & \text{if $\mathbb{P}(\Vert X_n\Vert>M_n)<\epsilon$}\\
M_{n}+1 & \text{if $\mathbb{P}(\Vert X_n\Vert>M_n)\geq\epsilon$}
\end{cases}
\end{equation}
First, we prove that $M_n\rightarrow \infty$. For $n\in\mathbb{N}$, we already have that, for $M_n$, there is some $k\geq n$ such that $\mathbb{P}(\Vert X_k\Vert>M_n)>\epsilon$. Then we have $M_{k+1}=M_k+1$. This proves the sequence diverges to infinity. 
Second, since for all $n$, there is some $k\geq n$ such that $\mathbb{P}(\Vert X_k\Vert>M_n)>\epsilon$ we note we can extract a subsequence $k_n$ such that $\mathbb{P}(\Vert X_{k_n} \Vert>M_{k_n})\geq\epsilon$. This contradicts that 

$\mathbb{P}(\Vert X_n\Vert>M_n)\rightarrow 0$ as $n\rightarrow \infty$ for every $M_n\rightarrow \infty$

