For the second question suppose that $L$ is a geometrically rigid Lie algebra,
i.e., the orbit $O(L)$ is open. The algebra $L$, and hence $O(L)$ is contained
in some irreducible component $\mathcal{C}$. Since a nonemty open subset in an
irreducible space is dense, the closure of $O(L)$ equals $\mathcal{C}$. The first question
follows similarly.
It might also be helpful for understanding to use a more intuitive definition of rigidity. Call $L$ formally rigid,
if $L$ does not admit any non-trivial formal deformation. Over fields of characteristic
zero, geometrical rigidity is equivalent to formal rigidity (otherwise this is not true).
Assume that we have two Lie algebras $L_0$ and $L$ of the same dimension over
a field of characteristic zero. Suppose that $L_0$ degenerates properly to $L$, i.e.,
$L$ lies in the boundary of the orbit $O(L_0)$. Then this degeneration directly gives a
non-trivial deformation of $L$, so $L$ cannot be formally rigid, and hence not
geometrically rigid.