If $A$ is an infinite set, prove that $A$ has a proper infinite subset I considered taking this contradictory approach: Let $A_1\in A$. Assume that the proper subset $A\smallsetminus\{A_1\}$ is finite such that there exists a bijection $f:A\smallsetminus\{A_1\} \to C_k$ with $C_k=\{1,2,3,\ldots,k\}$ with $k \in N$.
What I wanted to do next is somehow prove that since $A\smallsetminus\{A_1\}$ is finite, then $A$ is finite, contradicting the given and proving my assumption false. However I have no idea how to proceed from here, or even if what I've done so far is plausible. Any help please?

I must assume that you mean that there exists a subset $B\subsetneq A$ such that it is bijective with $A$. Because otherwise it would only be enough to remove a point or two and you would have what you want: let $x\in A$, then $B=A\setminus\{x\}\subsetneq A$. End.
Since $A$ is infinity, then there exists $f:\mathbb N\to A$ injective. We take $X=f(\mathbb N)\subset A$. then $X=\{a_1,a_2,\dots\}$ with $a_i\neq a_j$ if $i\neq j$. Let $B=(A\setminus X)\cup \{a_2,a_4,\dots,a_{2n},\dots\}\subset A$. Now let's define $g:A\to B$, given that $$f(a)=\left\{\begin{array}{ccl}a&,&a\in A\setminus X\\
a_{2n}&,&a=a_{n}\end{array}\right.$$
It is clearly surjective, let's see that $g$ is injective. We suppose that $g(y)=g(z)$.
If $y,z\in A\setminus X$, then $y=g(y)=g(z)=z$.
If $y,z\in X$, then $y=a_n$ and $z=a_m$ for some $n,m\in\mathbb N$. So $g(y)=g(z)\Rightarrow a_{2n}=a_{2m}\Rightarrow 2n=2m\Rightarrow n=m \Rightarrow a_n=a_m\Rightarrow y=z$.
If $y\in A\setminus X$ and $z\in X$, then $y=g(y)=g(z)=a_n$ for some $n$, so $y=a_n\in X$. Contradiction.
Hence $g$ is a biyection between $A$ and $B\subsetneq A$.
