What you need is that the intersection of two subalgebras $X_1$ and $X_2$ is also a subalgebra:
$X_1$ and $X_2$ will both contain $0$ and $1$ $\implies$ then so must their intersection.
If $X_1$ and $X_2$ both contain element $a$ then they will both contain element $\lnot a$ $\implies$ then so must their intersection.
Closure:
If both $a$ and $b$ are in $X_1$ and $X_2$ $\implies$ then $a \land b$ and $a \lor b$ must both be in $X_1$ and in $X_2$ and so also in their intersection.
If $a$ and $b$ are in one $X_i$ but both not in the other $\implies$ then both $a$ and $b$ are not in their intersection so we don't have to prove closure in this case.
If $a$ and $b$ are in one $X_i$ (say $X_1$) but only one of them (say $a$) is in the other $\implies$ then $a$ is in their intersection, but not $b$ so we don't have to prove closure with respect to $a$ and $b$ in this case.
We conclude that all the axioms are satisfied so the intersection of subalgebras is a subalgebra.
This means that the intersection of one or more subalgebras containing a subset $X$ is a subalgebra containing subset $X$.
Note (after collecting some Latex symbols).
Of course the above is not the shortest way to prove closure. That would be something like:
Let $a$ , $b$ $\in X_1 \cap X_2 \implies \\ a , b \in X_1 \land a , b \in X_2 \implies ( a \land b , a \lor b \in X_1 ) \land ( a \land b , a \lor b \in X_2 ) \implies \\ a \land b , a \lor b \in X_1 \cap X_2 $
