Prove: If $X$ and $Y$ are infinite, then $X\cup Y$ is infinite. Proof:
Suppose $X$ and $Y$ are any infinite sets.
Then 
$$ \exists f:X \rightarrow X \text{ such that}\; f \;\text{ is injective }\;\land\; f(X) \neq X,\;\;\text{and}$$ 
$$\exists g:Y \rightarrow Y \text{ such that}\;\; g \text{ is injective}\;\;\land\;\;g(Y) \neq Y.$$
I'm sure that it's simple, but I don't see what I should do after this.
 A: Since $X \subset X \cup Y$, the same injection $f$ demonstrating that $X$ is infinite can also be made to demonstrate that $X \cup Y$ is infinite. 
For ease of notation, suppose $X$ and $Y$ are disjoint. (We won't count duplicates in the union, so this is a fair assumption.) Define $h : X \cup Y \rightarrow X \cup Y$ by
$$
h(z) = 
\begin{cases}
f(z) & \text{ if } z \in X\\
z & \text { if } z \in Y
\end{cases}
$$
Since $f$ is an injection, so is $h$. Now, 
$$
h(X \cup Y) = h(X) \cup h(Y) = f(X) \cup Y \neq X \cup Y,$$ 
where the inequality comes from the fact that $f(X) \neq X$.

If the assumption that $X$ and $Y$ are disjoint makes you uncomfortable, you can do without it by replacing $Y$ with $Y \setminus X$ in the definition of $h$ and throughout the proof.
A: Assume it is not true. Then there is a bijection with $(1, 2, ...., n)$ for some $n$ that counts all elements of the union. Then it (over-)counts all elements of, say, $X$. Then, for some $m<n$, $(1, 2, ..., m)$ counts $X$, which is a contradiction.
