Find the cardinality of $\{f: \mathbb{N} \rightarrow \mathbb{P}(\mathbb{N}) \mid ( \forall n ) ( n \in f(n) ) \}$ 
Find the cardinality of $A=\{f: \mathbb{N} \rightarrow \mathbb{P}(\mathbb{N}) \mid (\forall n ) ( n \in f(n) ) \}$.

I know $A\subseteq \mathbb{P}(\mathbb{N})^{\mathbb{N}}$, so ${\#}A \leq \mathfrak{c}$.
I want to show the other inequality by defining a function $\phi: \mathbb{P}(\mathbb{N}) \rightarrow A$. If $K \subseteq \mathbb{N}$ then $\phi(K)=f_{K}$ where $f_{K}: \mathbb{N} \rightarrow \mathbb{P}(\mathbb{N})$ is defined by $$f_{n}=
\begin{cases}
K, &\text{if }n \in K \\
\{ n \}, &\text{if }n \notin K.
\end{cases}
$$
Is this correct? If so how can I prove this function is injective. If I prove that then I think I prove ${\#}A=\mathfrak{c}$.

$\phi$ is not one-to-one, but allmost: All singletons are mapped to the same function. For each $m,n$ we have
$$ \phi(\{m\})(n) = f_{\{m\}}(n) = \{n\} $$
so $\phi(\{m\}) = \phi(\{m'\})$, every $m, m'$.
But if $|K| \ge 2$, and $K \ne L$, choose $n \in K$, if $n \not\in L$, then 
$$ f_K(n) = K \ne \{n\} = f_L(n) $$
and if $n\in L$, then 
$$ f_K(n) = K \ne L = f_L(n) $$
Hence $f_K \ne f_L$. As there are only $\aleph_0$ singletons in $\mathbb P(\mathbb N)$, the set $B := \{ K \in \mathbb P(\mathbb N)\mid |K| \ge 2\}$ has cardinality $\mathfrak c$. As - we showed this above - $\phi|_B \colon B \to A$ is one-to-one, $\mathfrak c = |B| \le |A|$. 
So, by Schröder-Bernstein, $|A| = \mathfrak c$.
