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I was wondering if someone could give me an example of an infinite set and a bijection from the set to a proper subset of itself.

$$f: mathbb Nrightarrow 2mathbb N$$
$$n mapsto 2n$$

It's a bijection from the set of all natural numbers $1, 2, 3, cdots$ into its proper subset of all even natural numbers $2, 4, 6, cdots$.

This is easy and more general than you think.

If $f:A to A$ is one to one then $f:Ato f(A)subseteq A$ is onto and a bijection. If $f:A to A$ is not onto then $f:A to f(A)subsetneq A$ is a bijection to a proper subset.

So all we need is 1) An infinite set $A$. 2) An injective but not onto function from $Ato A$.

$f:mathbb R to mathbb R$ via $f(x) = e^x$ is one to one and will always have $e^x > 0$ and the image of $f$ is $(0,infty)$.

So $f:mathbb R to (0,infty)$ via $f(x) = e^x$ is an example.

But probably the very first one most of us have probably heard of is:

"You know, there are the same number of even numbers and numbers because for every number if you multiply it by $2$ you get an even number."

Most of us heard that in the first grade or so (and lost three weeks of sleep immediately afterward).

This is $f: mathbb N to $even integers$$ via $f(n) = 2n$ is such a bijection.

Less easy but maybe more fundamental and more to the point is:

we know $|mathbb Q| = |mathbb Z| = |mathbb N|$ so by definition we know there are bijections from $mathbb Q to mathbb Z$ and from $mathbb Q to mathbb N$ and from $mathbb Z to mathbb N$.

So we can say if $qin mathbb Q^+$ and $q = frac ab > 0$ in lowest terms and we write down all the pairs of integers in the following order $(1,1),(2,1),(1,2),(3,1),(2,2),(1,3), (4,1),(3,2),(2,3),(1,4),(5,1),(4,2),.....$ and we cross out all the pairs that are not relatively prime so we get $(1,1),(2,1),(1,2),(3,1),(1,3),(4,1),(3,2),(2,3),(1,4),(5,1),(1,5),(6,1)....$ and count where $(a,b)$ appears in the list... then that is a bijection from $mathbb Q^+ to mathbb N$.

One example was given in the comments. Here’s another:

$$n mapsto n+1$$

is a bijection from $0,1,2,...$ to $1,2,...$ $subsetneq $ $0,1,2,...$.