Suppose $(\Omega,\mathcal F,P)$ is a probability space with probability measure $P$, $x(t,\omega)$ a random variable dependent on $t\in[0,T]$. $\forall \omega\in\Omega,\,\lim_{t\rightarrow0}\,x(t,\omega)\not\rightarrow0$. Does there exist a sequence $(t_k)_{k=1}^\infty\rightarrow0$ and a random variable $y$ satisfying either of the following conditions?
$y(\omega)\ne0,\,\forall\omega\in\Omega$, such that for any $\omega\in\Omega$, $x(t_k,\omega)\rightarrow y(\omega)$.
$\exists A\subset\Omega$, $P(A)>0,\,\ni \big(y(\omega)\ne0 \wedge x(t_k,\omega)\rightarrow y(\omega)\,\forall \omega\in A\big)$.
No. For instance, let the probability space be Lebesgue measure on $[0,1]$ and let $x(t,\omega)=1$ if $t\in [1/2^{n+1},1/2^n]$ and the $n$th binary digit of $\omega$ is $0$ and $x(t,\omega)=2$ otherwise. For any particular sequence $(t_k)$ converging to $0$, you can choose some $\omega$ such that $x(t_k,\omega)=1$ for infinitely many $k$ and also $x(t_k,\omega)=2$ for infinitely many $k$ (just let the relevant binary digits of $\omega$ oscillate).
This is also a counterexample to your second condition. For any $(t_k)$, the set of $\omega$ such that $x(t_k,\omega)$ converges has measure $0$ (since it is the set of numbers such that some particular infinite sequence of binary digits is either eventually $0$ or eventually $1$).