If $\mathfrak F$ is a partition of a collection $\mathcal A$ then it is not hard to show that the equality $$ \tag{1}\label{eq: general associative law for intersection}\bigcap_{A\in\mathcal A}A=\bigcap_{\mathcal F\in\mathfrak F}\left(\bigcap_{F\in\mathcal F}F\right) $$ and the equality $$ \tag{2}\label{eq: general associative law for union}\bigcup_{A\in\mathcal A}A=\bigcup_{\mathcal F\in\mathfrak F}\left(\bigcup_{F\in\mathcal F}F\right) $$ hold.
Now if $L$, $M$ and $N$ are mutually distinct then the collections $$ \mathfrak U:=\big\{\{L\},\{M,N\}\big\}\quad\text{and}\quad\mathfrak V:=\big\{\{L,M\},\{N\}\big\} $$ are partition of $\{L,M,N\}$ so that by eq. \eqref{eq: general associative law for intersection} we argue the equality $$ \tag{3}\label{eq: associative law for intersection with distinct factors} L\cap(M\cap N)= \bigcap\big\{L,M\cap N\big\}= \bigcap\left\{L,\bigcap\big\{M,N\big\}\right\}=\\ \bigcap\left\{\bigcap\big\{L\big\},\bigcap\big\{M,N\big\}\right\}= \bigcap\left\{\bigcap\big\{L,M\big\},\bigcap\big\{N\big\}\right\}=\\ \bigcap\left\{\bigcap\big\{L,M\big\},N\right\}= \bigcap\big\{L\cap M,N\big\}= (L\cap M)\cap N $$ and by eq. \eqref{eq: general associative law for union} we argue the equality $$ \tag{4}\label{eq: associative law for union with distinct factors} L\cup(M\cup C)= \bigcup\big\{L,M\cup N\big\}= \bigcup\left\{L,\bigcup\big\{M,N\big\}\right\}=\\ \bigcup\left\{\bigcup\big\{L\big\},\bigcup\big\{M,N\big\}\right\}= \bigcup\left\{\bigcup\big\{L,M\big\},\bigcup\big\{N\big\}\right\}=\\ \bigcup\left\{\bigcup\big\{L,M\big\},N\right\}= \bigcup\big\{L\cup M,N\big\}= (L\cup M)\cup N $$
Now let be $U$ the set $$ U:=\big\{\emptyset\big\} $$ so that let's we put $$ V:=U\cup\big\{U\big\}\quad\text{and}\quad W:=V\cup\big\{V\big\} $$ So we observe that $U$, $V$ and $W$ are mutually distinct (indeed, for who know the natural numbers $U$ is $1$, $V$ is $2$ and $W$ is $3$) since otherwidse $\emptyset$ would not be empty; moreover, we observe that the inclusive chain $$ U\subseteq V\subseteq W $$ holds. Now a not empty (binary) cartesian product is equal to another not empty (binary) cartesian product if the first and the second factor of the first product are respectively equals to the first and second factor of the second product so that if $A$ $B$ and $C$ are not empty then $A\times U$, $B\times V$ and $C\times W$ are distinct and thus by \eqref{eq: associative law for intersection with distinct factors} the equality $$ \tag{5}(A\times U)\cap\big((B\times V)\cap(C\times W)\big)=\big((A\times U)\cap(B\times V)\big)\cap(C\times W) $$ holds so that by this formula we argue the equality $$ \big(A\cap(B\cap C)\big)\times U= \big(A\cap(B\cap C)\big)\times(U\cap V)=\\ (A\times U)\cap\big((B\cap C)\times V\big)= (A\times U)\cap\big((B\cap C)\times(V\cap W)\big)=\\ (A\times U)\cap\big((B\times V)\cap(C\times W)\big)= \big((A\times U)\cap(B\times V)\big)\cap(C\times W)=\\ \big((A\cap B)\times(U\cap V)\big)\cap(C\times W)= \big((A\cap B)\times U\big)\cap(C\times W)=\\ \big((A\cap B)\cap C)\times(U\cap W)= \big((A\cap B)\cap C\big)\times U $$ and thus by what above observed about equality of not empty (binary) cartesian products we conclude that the equality $$ \tag{6}\label{eq: associative law for binary intersection}A\cap(B\cap C)=(A\cap B)\cap C $$
Now we remember that if $H$ and $K$ a respectively two not empty sets of a set $M$ and a set $N$ then the equalities $$ H=\pi_M[H\times K]\quad\text{and}\quad K:=\pi_N[H\times K] $$ hold. So let's we put $$ X:=\bigcup\{A,B,C\}\quad\text{and}\quad Y:=\bigcup\{U,V,W\} $$ and thus let's we observe the inclusions $$ A,B,C\subseteq X\quad\text{and}\quad U,V,W\subseteq Y $$ hold. So if $A\times U$, $B\times V$ and $C\times W$ are mutually distinct then by eq. \eqref{eq: general associative law for union} the equality $$ \tag{7}(A\times V)\cup\big((B\times V)\cup(C\times W)\big)=\big((A\times V)\cup(B\times V)\big)\cup(C\times W) $$ holds and thus by what above observed and by "compatibility" of projections with respect union we argue the equality $$ \tag{8}\label{eq: associative law for binary union}A\cup(B\cup C)= \pi_X[A\times U]\cup\big(\pi_X[B\times V]\cup\pi_X[C\times W]\big)=\\ \pi_X[A\times U]\cup\pi_X\big[(B\times V)\cup(C\times W)\big]= \pi_X\Big[(A\times U)\cup\big((B\times V)\cup(C\times W)\big)\Big]=\\ \pi_X\Big[\big((A\times U)\cup(B\times V)\big)\cup(C\times W)\Big]= \pi_X\big[(A\times U)\cup(B\times V)\big]\cup\pi_X[C\times W]=\\ \big(\pi_X[A\times U]\cup\pi_X[B\times V]\big)\cup\pi_X[C\times W]= (A\cup B)\cup C $$
Finally, we observe that if $A$, $B$ and $C$ are empty then \eqref{eq: associative law for binary intersection} and eq. \eqref{eq: associative law for binary union} trivially hold.
So proving eq. \eqref{eq: general associative law for intersection} and eq. \eqref{eq: general associative law for union} I actually observed that it is not necessary for $\mathfrak F$ to be a partition but it is really necessary that $\mathcal A$ is union of $\mathfrak F$ so that in this way the proof of associative law for union and intersection would be more simple. However, I feel that by an algebraic point of view it would be better to require that $\mathfrak F$ is a partition so that I thought to put here a specific question where I first ask if I well proved associativity for intersection and union assuming $\mathfrak F$ is a partition and then where I ask some explanation about this hypothesis.