How to complete this proof? Union of a countably infinite set and a finite set is countably infinite Theorem. Let $X$ be a countably infinite set and $Y$ be a finite set. Then $X\cup Y$ is countably infinite.
Proof. Since $X$ is a countably infinite set, then there exists a bijection function $f\colon \mathbb{N}\longrightarrow X$ and since $Y$ is a finite set, then $X\cup Y$ is (why) is infinite. Therefore there exists (why) a bijection function $g\colon X\longrightarrow X\cup Y$ so $g\circ f\colon \mathbb{N}\longrightarrow X\cup Y$ is a 1-1 and surjective function. This means that $X\cup Y$ is countably infinite.
Could you please tell me the reason of the two WHYs?

The first “why” has no explanation; the fact that $X\cup Y$ is infinite stems from $X$ being infinite.
The second “why” is exactly the whole proof.
Since $Y$ is finite, also $Y'=Y\setminus X$ is finite. Note that $X\cup Y=X\cup Y'$ and $X\cap Y'=\emptyset$. Since $Y'$ is finite, there exist $d\in\mathbb{N}$ and a bijection
$$
h\colon\{0,1,\dots,d-1\}\to Y'
$$
Define
$$
g\colon\mathbb{N}\to X\cup Y'
$$
by
$$
g(n)=\begin{cases}
h(n) & \text{if $n<d$}\\[6px]
f(n-d) & \text{if $n\ge d$}
\end{cases}
$$
Can you prove $g$ is bijective?
(Note: if you're afraid about the case when $d=0$, that is, $Y'=\emptyset$, don't be; anyhow, in this case there's exactly nothing to prove.)
