A magma is a pair given $(X,\bot)$ given by a set $X$ and a function from $X\times X$ to $X$ so that we say that $Y\in\mathcal P(X)$ is closed under $\bot$ if the inclusion $$ \bot[Y\times Y]\subseteq Y $$ holds.
So we immediately observe that if $\mathcal U$ is a collection of closed set of $X$ under $\bot$ with not empty intersection then even its intersection $\bigcap\mathcal U$ is closed under $\bot$ since if $u_1$ and $u_2$ are in $\bigcap U$ then they are even in $U$ for any $U\in\mathcal U$ so that $u_1\bot u_2$ is in $U$ and thus trivially in $\bigcap U$.
Now $X$ is trivially closed with respect $\bot$ so that the collection $$ \mathcal S(\bot):=\{U\in\mathcal P(X):\bot[U\times U]\subseteq U\} $$ is not empty and thus for any $Y\in\mathcal P(X)$ the collection $$ \mathcal U_Y:=\{U\in\mathcal S(\bot):Y\subseteq U\} $$ is not empty and additionally if $Y$ is not empty and if any $U\in\mathcal U_Y$ contains $Y$ then their intersection $\langle Y\rangle$ is not empty so that it is the smallest closed set of $X$ under $\bot$ containing $Y$: in fact, what proved above implies that $\langle Y\rangle$ is closed and moreover if $U\in\mathcal P(X)$ is closed and contains $Y$ then $U$ is in $\mathcal U_Y$ so that the inclusion $$ \langle Y\rangle\subseteq U $$ holds.
First of all we observe that the operator $$ \langle\quad\rangle:\mathcal P(X)\ni Y\longrightarrow\langle Y\rangle\in\mathcal P(X) $$ is monotonous with respect the inclusion: in fact, if $U,V\in\mathcal P(X)$ are such that $$ U\subseteq V $$ then the inclusion $$ U\subseteq V\subseteq\langle V\rangle $$ holds so that what above showed implies that $$ \langle U\rangle\subseteq\langle V\rangle $$ Moreover if $Y\in\mathcal P(X)$ is closed then it is in $\mathcal U_Y$ so that the inclusion $$ \langle Y\rangle\subseteq Y $$ holds and thus if trivially even the inclusion $$ Y\subseteq\langle Y\rangle $$ holds then we finally conclude that $$ Y=\langle Y\rangle $$ which additionally implies that $\langle\quad\rangle$ is idempotent.
Now let be $\mathfrak V$ a partition of $\mathcal Y\subseteq\mathcal P(X)$ so that let's we prove the equality $$ \tag{0}\label{0}\Biggl\langle\bigcup_{Y\in\mathcal Y} Y\Biggl\rangle = \Biggl\langle\bigcup_{\mathcal V\in\mathfrak V}\biggl\langle\bigcup_{V\in\mathcal V} V\biggl\rangle\Biggl\rangle $$ holds.
So if $\mathfrak V$ partition $\mathcal Y$ then the equality $$ \bigcup_{Y\in\mathcal Y}Y = \bigcup_{\mathcal V\in\mathfrak V}\Biggl(\bigcup_{V\in\mathcal V}V\Biggl) $$ holds and so for any $V\in\mathfrak V$ the inclusion $$ \bigcup_{V\in\mathcal V}V\subseteq\bigcup_{Y\in\mathcal Y}Y \subseteq \Biggl\langle\bigcup_{Y\in\mathcal Y}Y\Biggl\rangle $$ holds; but we know that $\biggl\langle\bigcup_{V\in\mathcal V}V\biggl\rangle$ is the smallest closed set containing $\bigcup_{V\in\mathcal V}$ so that the last inclusion implies that $$ \bigcup_{V\in\mathcal V}\biggl\langle\bigcup_{V\in\mathcal V}\biggl\rangle\subseteq \Biggl\langle\bigcup_{Y\in\mathcal Y}\Biggl\rangle $$ and thus reapplying the argument finally we conclude that $$ \tag{1}\label{1}\Biggl\langle\bigcup_{\mathcal V\in\mathfrak V}\biggl\langle\bigcup_{V\in\mathcal V}V\biggl\rangle\Biggl\rangle \subseteq \Biggl\langle\bigcup_{Y\in\mathcal Y}Y\Biggl\rangle $$ However for any $Y\in\mathcal Y$ there exists $\mathcal V_Y\in\mathfrak V$ such that $$ Y\subseteq\biggl\langle\bigcup_{V_Y\in\mathcal V_Y}V_Y\biggl\rangle $$ so that taking the union over $\mathcal Y$ we conclude that $$ \bigcup_{Y\in\mathcal Y}Y \subseteq \bigcup_{Y\in\mathcal Y}\biggl\langle\bigcup_{V_Y\in\mathcal V_Y}V_Y\biggl\rangle \subseteq \bigcup_{\mathcal V\in\mathfrak V}\biggl\langle\bigcup_{V\in\mathcal V}V\biggl\rangle\Biggl\rangle \subseteq \Biggl\langle\bigcup_{\mathcal V\in\mathfrak V}\biggl\langle\bigcup_{V\in\mathcal V}V\biggl\rangle\Biggl\rangle\Biggl\rangle $$ which implies that $$ \tag{2}\label{2}\Biggl\langle\bigcup_{Y\in\mathcal Y}Y\Biggl\rangle \subseteq \Biggl\langle\bigcup_{\mathcal V\in\mathfrak V}\biggl\langle\bigcup_{V\in\mathcal V}V\biggl\rangle\Biggl\rangle $$ So by \eqref{1} and by \eqref{2} we conclude that the equality \eqref{0} holds.
Now I am quite sure what I have done above is quite correct but I deliberated to do it because e.g. I know that \eqref{0} is surely corrected for subgroup as here the professor Magidin said lefting for exercise the proof: so surely I ask if I well proved all things I did until now. However now I asked to me if additionally the equality
$$ \tag{3}\label{3} \biggl\langle\bigcap_{Y\in\mathcal Y}Y\biggl\rangle = \bigcap_{Y\in\mathcal Y}\langle Y\rangle $$ holds.
In fact, I observed that for any $Y\in\mathcal Y$ the inclusion $$ \bigcap_{Y\in\mathcal Y} Y \subseteq Y \subseteq \langle Y\rangle $$ holds so that even the inclusion $$ \bigcap_{Y\in\mathcal Y} Y \subseteq \bigcap_{Y\in\mathcal Y}\langle Y\rangle $$ but by above proved we know that $\bigcap_{Y\in\mathcal Y}\langle Y\rangle$ is closed so that the inclusion $$ \tag{4}\label{4} \biggl\langle\bigcap_{Y\in\mathcal Y}Y\biggl\rangle \subseteq \bigcap_{Y\in\mathcal Y}\langle Y\rangle $$ holds.
So I was not able to prove the reverse of \eqref{4} so that I ask if it to prove it or disprove it with a counterexample. So could someone help me, please?
Your proof of $(1),(2),(0)$ is ok.
Now, in $X=\{0,1\}$ with $\forall x,y\in X\quad x\bot y=0,$ define $Y_k=\{k\}$ ($\forall k\in X$). Then, $$\langle Y_0\cap Y_1\rangle=\langle\varnothing\rangle=\varnothing,$$ whereas $$\langle Y_0\rangle\cap\langle Y_1\rangle=\{0\}\cap X=\{0\}.$$