Trying to prove $\liminf (A_n) \subseteq \limsup (A_n)$ I am trying to show that for a sequence of sets $(A_n)$
$$ \liminf (A_n) \subseteq \limsup (A_n)$$
I can see why this is true intuitively as follows -
$x \in \liminf (A_n)$
$\implies x$ is an element of all but finitely many of the $A_n$'s
$\implies x$ is an element of infinitely many of the $A_n$'s
$\implies x \in \limsup (A_n)$

However I want to show it formally using set notation. Here is what I have so far.
$x \in \liminf(A_n)$
$\implies x \in \bigcup_n \bigcap_{k \ge n} A_k$
$\implies x \in \bigcap_{k \ge n_0} A_k$ for some $n_0$
$\implies x \in \big(\bigcup_{1 \le j < n_0} A_j\big) \bigcup \big(\bigcap_{k \ge n_0} A_k \big)$ for some $n_0$
$\implies x \in \big(\bigcup_{1 \le j < n_0} A_j\big) \bigcup \big(\bigcup_{k \ge n_0} A_k \big)$ for some $n_0$
But I am not sure how to proceed from here. It seems like the $n_0$ is now 'redundant' as we have a statement saying that $x$ is an element least one $A_k$ whether that $A_k$ is 'before, equal to or after' $A_{n_0}$ in the sequence $(A_n)$. But this leads to
$\implies x \in \bigcup_n A_n$
which just says $x$ is an element of at least one of the infinitely many $A_n$, whereas I need to show that $x$ occurs infinitely often.
Where am I going wrong?

Let us start from the line $x\in \bigcap_{k\geq n_0} A_k$. This implies that $x\in A_k$ for all $k\geq n_0$ and hence $x\in \bigcup_{k\geq n} A_k$ for all $n\geq 1$. Therefore also $x\in\limsup A_n$.
