Aussume P != NP, are these assertions valid? Assume $P \ne NP$, and $A$ is a problem in $P$ and $B$ is a problem which is $NP-complete$.
Are the following assertions valid?

$A \le_{P} B$
$B \le_{P} A$


My approach:

$B \le_{P} A$ isn't valid, because if $B$ is reducible in poly-time to $A$, so $B$ has to be also in $P$. But thats not possible because $P \ne NP$.
$A \le_{P} B$. Is it possible to reduce a problem in $P$ to a problem in $NP$ ? If it is, then the assertion is valid.

Question: Is my approch correct or am I missing some important point?
Regarding 1: yes, since $B$ is $\mathsf{NP}$-complete it must be $\mathsf{NP}$-hard, i.e., for ever every problem $C$ in $\mathsf{NP}$ we have $C \le_P B$. Since $A \in \mathsf{P} \subseteq \mathsf{NP}$, we can pick $C=A$.
This is true regardless of whether $\mathsf{P}=\mathsf{NP}$.
Notice, however, that we are using the fact that $B$ is $\mathsf{NP}$-complete. In general it is not possible to reduce an arbitrary problem in $\mathsf{P}$ to an arbitrary problem in $\mathsf{NP}$.
Regarding 2: no. Here your reasoning is correct, $B \le_p A$ would imply $\mathsf{P} = \mathsf{NP}$.