Inclusions of constructible sets are *not* definable 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 $L$.
 A: 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 $L$.  
A: The trick to prove the powerset axiom in $L$ is to use condensation. Here is how one proves that $P(\omega) \cap L \in L$.
We show that $P(\omega) \cap L \subseteq L_{\omega_1}$ (and this is the best possible bound). If $x \subseteq \omega$ is constructible then $x \in L_\alpha$ for some ordinal $\alpha$. Let $N$ be a countable elementary submodel of $L_\alpha$ such that $x \in N$. By the Condensation Lemma, the Mostowski collapse of $N$ is some $L_\delta$, where $\delta < \omega_1$ since $N$ is countable. Since every element of $\omega$ collapses to itself, $x$ collapses to $x$ and hence $x \in L_{\delta}$. Therefore $x \in L_{\omega_1}$ as claimed.
For subsets of a general infinite constructible set $a$, one needs to use an elementary submodel $N$ which contains the transitive closure of $a$. This way, every element of $a$ collapses to itself and hence every subset of $a$ in $N$ gets collapsed to itself. One can always find such a $N$ with size equal to the size of the transitive closure of $a$. Therefore $P(a) \cap L \subseteq L_{\kappa^+}$, where $\kappa$ is the size of the transitive closure of $a$.
