Is $\mathbb{N}\cup\{a\}$, for some $a\not\in\mathbb{N}$ countable or uncountable?
$\mathbf{Attempt: }$ It is true that a set is countable if there exists an injective function $f : S → N$ from $S$ to the natural numbers $\mathbb{N}$. Since $\mathbb{N}\cup\{a\}$ has cardinality $|\mathbb{N}|+1$, $\mathbb{N}\cup\{a\}$ is not countable. However, there is a surjective mapping from $\mathbb{N}\cup\{a\}\to\mathbb{N}$. A set $X$ is uncountably infinite if $X$ is nonempty and there is no surjective function from the natural numbers to $X$. $\mathbb{N}\cup\{a\}$ is nonempty, and there is a surjective mapping from $\mathbb{N}\cup\{a\}\to\mathbb{N}$. But we had said that it is not countably infinite. This seems to be a contradiction, so $\mathbb{N}\cup\{a\}$ is countably infinite.
Hint:
Let $f: \Bbb N \rightarrow \Bbb N \cup \{a\}$ be defined thus. $$f(1) = a \; \text{and } \; f(x) = x - 1 \; \text{for every member in $\Bbb N \setminus \{1\}$} $$
You can show that this is a bijection.