Let $F$ be the set of all functions $f: \mathbb{Z^+} \to \mathbb{Z^+}$. Is $F$ countable or uncountable?
Define $K: \mathcal{P}(\mathbb{Z^+}) \to \{0,1\}^{\mathbb{Z^+}}$ for $U \subseteq \mathbb{Z^+}$ by letting $K(U)=\chi_{U}:\mathbb{Z^+} \to \{0,1\}$,
where $\chi_{U}(a)$=$\left\{ \begin{array}{lr} 1 & if \hspace{2mm} a \in U\\ 0 & if \hspace{2mm} a \notin U \\ \end{array} \right.$
Suppose $U,V \subseteq \mathbb{Z^+}$ s.t. $K(U)=K(V)$. Then $\chi_{U}=\chi_{V}$. Because $U=V \iff \chi_{U}=\chi_{V}$, we have $K(U)=K(V) \Rightarrow \chi_{U}=\chi_{V} \Rightarrow U=V$. This means that $K$ is injective and thus $|\mathcal{P}(\mathbb{Z^+})| \le |\{0,1\}^{\mathbb{Z^+}}|$.
By Cantor's Theorem, $\mathcal{P}(\mathbb{Z^+})$ is uncountable, and since $\exists$ injection $K: \mathcal{P}(\mathbb{Z^+}) \to \{0,1\}^{\mathbb{Z^+}} \Rightarrow \{0,1\}^{\mathbb{Z^+}}$ is uncountable.
As $|\mathbb{Z^+}^\mathbb{Z^+}| \ge |\{0,1\}^{\mathbb{Z^+}}| \ge |\mathcal{P}(\mathbb{Z^+})|=2^{|\mathbb{Z^+}|} \Rightarrow$ $F$ is uncountable.
I'm not sure whether this would be an appropriate method to prove this.
This set would also be a nice cacndidate for Cantor'd diagonal argument directly: Suppose $F$ is countable, i.e., there is an enumeration $F=\{f_1,f_2,f_3,\ldots\}=\{\,f_n\mid n\in\Bbb N\,\}$. Define $$ g(n)=f_n(n)+1$$ Then $g\in F$, gut $g\ne f_n$ for all $n$.