Proof that an injective system with $i\notin\textrm{ran}{f}$ has a Peano subsystem. I follow this link, in particular Exercise 2 at the bottom of page 3.
Def 1. A System is a tripple $(X,i,f)$, where $X$ is a set, $i$ is called initial element, and $f$ is a function $X\to X$.
Def 2. $S_2$ is a subsystem of $S_1$ if (a) $S_2\subset S_1$, (b) $i_1=i_2$, and (c) $\forall x\in S_2[f_1(x)=f_2(x)]$. proper subsystem implies $S_1\ne S_2$.
Def 3. A Peano's system is a system $(X,i,f)$ such that (1) $f$ is injective, (2) $i\notin \textrm{ran}{f}$, and (3) $(X,i,f)$ has no proper subsystem.
Lemma Given a system $(X,i,f)$ such that (1) $f$ is injective, and (2) $i\notin \textrm{ran}{f}$, then $(X,i,f)$ has a Peano's subsystem.
Proof
A subsystem has the same initial element and the same function. We define a set using the axiom of separation:
$$
\hat{X}:=\{x\in X: x=i\lor \exists x'[x'\in \hat{X}\land x=f(x')]\}.   
\quad\quad(*)
$$
Now we prove that $(\hat{X},i,f)$ is a Peano's system. (1) $\hat{X}\subset X$ thus $f\mid_{\hat{X}}$ is injective. (2) Since $\textrm{ran}{f\mid_{\hat{X}}}\subset\textrm{ran}{f}$, follows that $i\notin \textrm{ran}{f\mid_{\hat{X}}}$. (3) Now suppose that $(\hat{X},i,f)$ has a proper subsystem $(\hat{X}',i,f)$. Then $\hat{X}'\subsetneq \hat{X}$ and there is an $x\in \hat{X}$ such that $x\notin \hat{X}'$. From $(*)$ we know that either $x=i$ or $\exists x'(x=f(x'))$. $x=i$ is not possible because, being $(\hat{X}',i,f)$ a subsystem, $i\in \hat{X}'$. We consider $x\ne i$, then from (*) we know that $x'\in \hat{X}$. Two cases again: $x'\in\hat{X}'$ leads to a contradiction since $f$ is defined on $x'$, and so on until reaching $i$ and getting a contradiction.

How can I conclude the proof formally (case $x\ne i$)? I assume that the exercise should be "easy" since the proofs that follow are not very difficult. 
  The proof should not include external results or the use inductive concept that is introduced later.

 A: I don't think your definition of $\hat{X}$ works, which is where the problems are coming from. Here, let me help you rebuild your approach a bit.
Every system $X$ has a smallest subsystem $\hat{X} \subseteq X$, which can be defined in a number of ways. One way is to define $\hat{X}$ as the intersection of all the subsystems (and to check that being a subsystem is closed under intersection).
After this, you should check that $\hat{X}$ has no proper subsystems. I haven't thought about it too carefully, but I think the relevant fact is:

Lemma. Let $\varphi : X \rightarrow Y$ denote an injective morphism of systems. Then the inverse-image function $\varphi^* : \mathrm{Sub}(Y) \rightarrow \mathrm{Sub}(X)$ is surjective. When restricted to the image of the direct image map $\varphi_* : \mathrm{Sub}(X) \rightarrow \mathrm{Sub}(Y)$, it's bijective.

Once you've proved this, define $\varphi : \hat{X} \rightarrow X$ to be the inclusion; you should be able to use the above lemma to ensure there's no extra subsystems lurking around.
Now define that a system $X$ is injective iff $f_X$ is injective, and weakly acyclic iff $0_X \notin \mathrm{img}(f_X)$. All that remains to be done is to prove that these conditions are preserved under the taking of subsystems. This is kind of obvious, so once you get this point, you're basically done.
Some further terms to look up:


*

*Monounary algebra

*Pointed monounary algebra

*NNO (natural numbers object)

