3. Let integer $n \geqslant 2, X$ be an $n$-element set. For a sequence of non-empty subsets $A_{1}, A_{2}, \cdots, A_{k}$ of $X$, if $A_{1} \cup A_{2} \cup \cdots \cup A_{k}$ is a proper subset of $X$, and no element of $X$ is in exactly one $A_{i}$, then the sequence $A_{1}, A_{2}, \cdots, A_{k}$ is called "tight". Let $A_{1}, A_{2}, \cdots, A_{k}$ be a sequence of non-empty proper subsets of $X$, such that the sequence and any of its subsequences are not tight. Find the maximum value of $k$.

Note: For a subset $A$ of $X$, if $A \neq X$, and all sets in the sequence are distinct, then $A$ is called a proper subset of $X$.
3. $k_{\max }=2 n-2$.

Let $X=\{1,2, \cdots, n\}$.
On one hand, take $B_{i}=\{1,2, \cdots, i\}$,
$C_{i}=\{i+1, i+2, \cdots, n\}(i=1,2, \cdots, n-1)$,
$A_{1}, A_{2}, \cdots, A_{k}=B_{1}, B_{2}, \cdots, B_{n-1}$,
$C_{1}, C_{2}, \cdots, C_{n-1}$.
Thus, some $A_{i}$ are tight.
If $B_{i}$ is selected, then $C_{j}(j \leqslant i)$ cannot be selected. Otherwise, $B_{i} \cup C_{j}=X$. Take the maximum $i$. Thus, $i$ cannot be in $B_{k}(kj)$. Hence, $j+1$ appears only once.
On the other hand, we need to prove $k \leqslant 2 n-2$.
We use induction on $n$.
For $n=2$, $A_{i}=\{1\}$ or $\{2\}$. Thus, $k \leqslant 2$.
Assume the statement holds for an $(n-1)$-element set $X^{\prime}$. Consider an $n$-element set $X$.
Without loss of generality, assume 1 appears the least number of times in $A_{i}$.
Let $1 \in A_{i}(1 \leqslant i \leqslant t), 1 \notin A_{i}(t+1 \leqslant i \leqslant n)$.
Then $\underset{t+1 \leqslant i \leqslant k}{\cup} A_{i} \neq X$.
Since $A_{i}(t+1 \leqslant i \leqslant k)$ are not tight, there exists an element $i$ that appears exactly once in some $A_{j}(t+1 \leqslant i \leqslant k)$.
Without loss of generality, let $i=2, j=k$.
Since 1 appears the least number of times, which is $t$ times, and 2 appears exactly once in $A_{i}(t+1 \leqslant i \leqslant k)$, then 2 appears at least $t-1$ times in $A_{i}(t+1 \leqslant i \leqslant t)$.

Let $2 \in A_{i}(2 \leqslant i \leqslant t)$. Then in $A_{i}(2 \leqslant i \leqslant k-1)$,
$1 \in A_{l} \Leftrightarrow 2 \in A_{l}(2 \leqslant l \leqslant k-1)$.
Thus, we can merge 1 and 2 into a single element $1^{\prime}$,
$$
X^{\prime}=\left\{1^{\prime}, 3, \cdots, n\right\}.
$$

Therefore, $A_{i}^{\prime}(2 \leqslant i \leqslant k-1)$ and $X^{\prime}$ satisfy the induction hypothesis. Hence, $k-2 \leqslant 2 n-4 \Rightarrow k \leqslant 2 n-2$.