It’s not quite right: you’ve actually assumed that $x\in\limsup_kE_k$, so the argument is circular. Recall that
$$\liminf_{k\in\Bbb N}E_k=\bigcup_{n\ge 0}\bigcap_{k\ge n}E_k\;;$$
this means that if $x\in\liminf_{k\in\Bbb N}E_k$, then there is some $n_0\in\Bbb N$ such that $x\in\bigcap\limits_{k\ge n_0}E_k$.
Since $$\limsup_{k\in\Bbb N}E_k=\bigcap_{n\ge 0}\bigcup_{k\ge n}E_k\;,$$ you have to use this somehow to show that $x\in\bigcup\limits_{k\ge n}E_k$ for every $n\in\Bbb N$. That’s not actually very hard: we know that $x\in E_k$ for each $k\ge n_0$, so if we just take $k=\max\{n,n_0\}$ then ... ?
Note that if $A_n$ is any family of sets, then
$$ \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=0}^{n-1} A_k\bigr)$$
where the summands on the right-hand side are disjoint, and each of them is constructed from finitely many of the $A_i$s by a sequence of complements and finite unions.
So if you choose you can restrict yourself to requiring countable unions of disjoint set, if you also require that the algebra is closed under finite unions of arbitrary sets.
A Dynkin system that is not a $\sigma$-algebra:
If you don't require arbitrary finite unions, what you get is not necessarily a $\sigma$-algebra. Consider for example the system of subsets of $\mathbb R$ consisting of $\varnothing$, $\{0,x\}$ for every $x\ne 0$, and the complements of these sets. It is closed under your proposed axioms, because the only nontrivial disjoint union there is to take is $\{0,x\} \cup \{0,x\}^\complement = \mathbb R$.
A small finite example is
$$ \bigl\{\varnothing, \{0,2\}, \{0,3\}, \{1,2\}, \{1, 3\}, \{0,1,2,3\} \bigr\} $$
Best Answer
Clearly a $\sigma$-algebra, $\mathscr{A}$ on a set $X$ is by definition closed under "all kinds of" countable unions of sets that belongs to $\mathscr{A}$.
Now suppose $\mathscr{A}$ is an algebra on $X$, which is closed under countable increasing unions.
Pick an arbitrary sequence of sets $\left\{A_{n}\right\}_{n=1}^{\infty}\subset \mathscr{A}$ and consider the sequence $\left\{B_{k}\right\}_{k=1}^{\infty}$, defined by $$B_{k}= \bigcup_{n=1}^{k}A_{n}$$
Now indeed $B_{k}\subseteq B_{k+1}$ and $B_{k}\in \mathscr{A}$, for all $k\geq1$, since $\mathscr{A}$ is an algebra. Moreover $$\bigcup_{n=1}^{\infty}A_{n}=\bigcup_{k=1}^{\infty}B_{k}\in \mathscr{A}$$
which finishes the proof.