Let $L$ be the constructible universe and $x,y \in L$ such that $x \subseteq y$. Is then $x \in D(y)$, i.e. $x$ a definable subset of $y$?


If this is not true, do we at least have the following: If $x,y \in L$, then $\rho(x) \leq \rho(y)$? Here $\rho$ denotes the $L$-rank.


And if this is also false, how do you prove, that for $x \in L$, we have $L \cap P(x) \in L$? So I need this to prove the power set axiom in $L$.



Inclusions of constructible sets are *not* definable
If you just want to prove that $L$ satisfies the power set axiom, you don't need condensation; Kunen's comment is correct. Given $x\in L$, to show that $L\cap P(x)\in L$, first consider an arbitrary $y\in P(x)\cap L$ and observe that, since $y\in L$, there is a first ordinal $\alpha(y)$ such that $y\in L\_{\alpha(y)}$. By replacement (in the real world), there is an ordinal $\beta$ that is larger than all these $\alpha(y)$ for all $y\in P(x)\cap L$. So $P(x)\cap L\subseteq L\_\beta$. But then $P(x)\cap L$ is first-order definable over $L\_\beta$, by the definition $\{y\in L\_\beta:L\_\beta\models y\subseteq x\}$, and is therefore an element of $L\_{\beta+1}$. 


Note that this proof, unlike the one given by Francois, tells you nothing about how far you must go in the $L$ hierarchy before you find $P(x)\cap L$; it provides no information about the size of $\beta$. To get such information, condensation is needed.


Concerning the first two parts of the original question, the answer is no for both in general. Take $y$ to be $\omega$ and note that it has only countably many definable subsets and only countably many subsets of lower or equal $L$-rank, but it may have (and will have if, for example, $V=L$) uncountably many subsets in $L$. 


