Question about solution to showing $\mathbb{P}(\lim_{s\to t} X_s \;\text{exists})=0$ if $X_t$ are independent $N(0,t)$. This is a problem from Rene Schilling's Brownian Motion.
Let $(X_t)_{t\ge 0}$ be a stochastic process such that $X_t$ are independent  $N(0,t)$ random variables. Show that $\mathbb{P}(\lim_{s\to t} X_s \;\text{exists})=0$ for every $t>0$. 
Below is the solution to the problem. In the solution, I'm not sure how we get a uniform q by step 1. So in step 1  we find out that for any sequence $t_n \to t$, we have $P(\lim X_{t_n}\; \text{exists}) <1$. But $P(\lim X_{t_n}\; \text{exists})$  depends on the particular sequence $t_n$ chosen. So how do we get a uniform $q \in (0,1)$ that is $q > P(\lim X_{t_n}\; \text{exists})$ for all $t_n$?


 A: I think there are numerous problems in the solution, though the idea is somewhat clear.
Consider the following proof instead.
Take $t>0$, and fix a sequence $t_n\to t$ such that it is strictly decreasing, so that, in particular, each $t_n > t$. Consider the event $$E:=\{\omega\mid X_{t_n}(\omega)\mbox{ converges}\}.$$ Fix $\epsilon > 0$. Consider further the event $$E_n:=\{\omega\mid \forall m > n: |X_{t_m}(\omega)-X_{t_n}(\omega)|<\epsilon\}.$$ Since for every $m > n$, $t<t_m<t_n$, $X_{t_m}$ is independent of $X_{t_n}$ having variance at least $t$, and the normal distribution $N(0,t_m)$ is unimodal centered at $0$, hence, $$\begin{align*}\mathbb{P}\{|X_{t_m}-X_{t_n}|<\epsilon\} & \le\mathbb{P}\{|X_{t_m}|<\epsilon\} \\ & \le \mathbb{P}\{|N(0,t)|<\epsilon\} \\ & =c \\ & <1\mbox{ (as }t>0\mbox{)}.\end{align*}$$ From here, you conclude that $$\mathbb{P}\{E_n\}=0,$$and, since $$E\subset \cup_n E_n,$$we have $$\mathbb{P}\{E\}=0.$$ Then, the result follows, as $$\{\omega\mid \exists\lim X_s(\omega)\mbox{ as }s\to t\}\subset E.$$
P.S. There was a question, how to show that $\mathbb{P}\{E_n\}=0$, because $X_{t_m}-X_{t_n}$ are not independent.
Well, why this is true, it does not really matter in this case. The inequalities showing that $\mathbb{P}\{|X_{t_m}-X_{t_n}|<\epsilon\}\le c<1$ mean that this holds conditional on any value of $X_{t_n}$, and, hence, on any event.
Formally, define $$E_{n,k}=\{\omega\mid |X_{t_m}-X_{t_n}|<\epsilon\mbox{ for }m=n+1,\dotsc,n+k\}.$$ Further, define $$F_{n,k}=\{\omega\mid |X_{t_{n+k}}-X_{t_n}|<\epsilon\}.$$ Then, $$\begin{align*}\mathbb{P}\{E_{n,k+1}\} & =\mathbb{P}\{E_{n,k}\cap F_{n,k+1}\} \\ & =\mathbb{P}\{F_{n,k+1}\mid E_{n,k}\}\mathbb{P}\{E_{n,k}\} \\ & \le c \cdot \mathbb{P}\{E_{n,k}\}\end{align*}$$by the above, and $$E_n=\cap_k E_{n,k}.$$
