UPDATED: If $f(x + y) \leq yf(x) + f(f(x))$ for all real numbers $x$ and $y$, prove that $f(0) = 0.$ $\large\text{UPDATED:}$ (with completely correct arguments)

Let $f : \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies 
  $$f(x + y) \leq yf(x) + f(f(x))$$ for all real numbers $x$ and $y$.  Prove that $f(x) = 0$ for all $x ≤ 0$.  (IMO $2011$)

The purpose of my question is only proof verification. (not knowing the correct solution)
Here, I focus only on the case of $f(0) = 0.$ Because this is the main part of the problem and this is very easy to show that, $f(0) = 0$ follows $f(x) = 0$ for all $x ≤ 0.$ I want to prove only the $f(0)=0$.
Here are my steps:

Case $1.$ $f(0)\in \mathbb R^+$
We have,
$$f(0)\leq-xf(x)+f(f(x))$$
$$f(x)\leq xf(0)+f(f(0))$$
Applying $x \longrightarrow -\infty$ we get from $f(x)\leq xf(0)+f(f(0))$, $\lim_{x\to -\infty}f(x) = -\infty $.  
Then applying again $x \longrightarrow -\infty$, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0) \longrightarrow-\infty \not \in\mathbb R^+$, which gives a contradiction.

Case $2.$ $f(0)<0$ (with the wrong argument, e.g. $\lambda=0$)
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Case $2.$ $f(0) \in \mathbb {R^-}$ (with the correct argument)
We have, from $f(x)\leq xf(0)+f(f(0))$ we deduce $\lim_{x\to +\infty}f(x)=-\infty.$ From $f(x + y) \leq yf(x) + f(f(x))$ we have:
$\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$. 
Then applying $x=f(0)$, we get $f(f(0))\leq 0$, which imply $f(x)\leq 0$, which gives $f(f(x))\leq 0$. In this case, we have $f(x)<0.$ Because, if $f(x)=0$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$, which gives a contradiction. 
  Then, from  $f(x + y) \leq yf(x) + f(f(x))$  we have:
$f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(x) \leq (x-y)f(y)+f(f(y))\Longrightarrow 0\leq(f(y)-y)f(y) \Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x $
Applying $x\to-\infty$ from $f(0)\leq-xf(x)+f(f(x))$, we get $f(0)\longrightarrow -\infty \not \in \mathbb{R^-}$, which gives again a contradiction.
So, we can deduce that $f(0)=0$.
Q.E.D.

Can you verify the new solution?
I just want to make sure that I got $ f (0) = 0 $ correctly.
Thank you!
 A: I will do some comments on your redaction.
Case 1. $f(0)>0$

We have,
$$f(0)\leq-xf(x)+f(f(x))$$
$$f(x)\leq xf(0)+f(f(0))$$
Let $x\to -\infty$ we get from $f(x)\leq xf(0)+f(f(0)) \Longrightarrow \lim_{x\to -\infty}f(x)=-\infty$.

For a good redaction, don't mix $\Rightarrow$ with a french sentence.

Then applying $\lim_{x\to -\infty}f(x)=-\infty$, from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0) \longrightarrow-\infty.$ So, this is a contradiction.

Edit : OK. Precise clearly that the two terms tends to $- \infty$ to avoid fastidious verifications to the reader.

Case 2. $f(0)<0$
We have, from $f(x)\leq xf(0)+f(f(0))$ we deduce $\lim_{x\to +\infty}f(x)=-\infty.$ Suppose that, $\lim_{x\to -\infty}f(x)=+\infty.$ Applying $x\to-\infty$ from $f(x-1) ≤ -f(x) + f(f(x))$ we have  $\lim_{x\to -\infty}f(x+(-1)) \longrightarrow +\infty$.

You meant : $\lim_{x\to -\infty}f(x+(-1)) \rightarrow +\infty$ according to your assumption.

But, $\lim_{x\to -\infty} (-f(x) + f(f(x)))=-\infty$. According our assumption, we applied $\lim_{x\to +\infty}f(x)=-\infty.$ So, this is a contradiction.

Ok since $\lim_{x\to -\infty} f(f(x)) = - \infty$.

Suppose that, $\lim \inf_{x\to -\infty}f(x)=a$ and $\lim \sup_{x\to -\infty}f(x)=b$, where $a,b\in\mathbb{R}$

Ok. (A priori, $a, b \in \mathbb{R} \cup \{ - \infty \}$ but you deal with this after ) EDIT : to be more precise, $a \in \mathbb{R} \cup \{ - \infty \}$ and $b\in \mathbb{R} \cup \{ - \infty, +\infty \} $ ; you forgot the case $b = +\infty$ in your reasoning.

and for any $\lambda \in [a,b]$ we have $\lambda\leq y\lambda+f(\lambda)$.

This argument is interesting but problematic. I believe you have taken a sequence $x_n \rightarrow - \infty$ such that $f(x_n)$ tends to $\lambda$. This kind of argument is possible only if $f$ is supposed continuous (intuitively its graphe oscillates continuously between a and b). Furthermore, you cannot have a control on $f(x_n +y)$ while doing this (it might be improved by replacing $y$ by $y_n$). Finally, since again f is not supposed continuous, the behaviour of $f(f(x_n))$ might be chaotic and not converge at all to $f(\lambda)$.
If $f$ is supposed continuous, it is possible to make a (rigorous) proof. (X)
[[ EDIT : I said you needed the continuity for the first step because you have taken "any $\lambda \in [a, b]$". I think things will be clearer if I present the argument.
If you have continuity. You have two sequences $(a_n)$ and $(b_n)$ tending to $-\infty$ such that $$\lim_{n\rightarrow \infty} f(a_n) = a, \lim_{n \rightarrow \infty} f(b_n) = b$$
By the intermediate value theorem, you can find a sequence $(x_n)$ tending to $-\infty$, such that $f(x_n) \rightarrow \lambda$, and also (a little more technical) a sequence $(y_n)$ with $sup (y_n) = +\infty$, $inf (y_n) = -\infty$ such that $f(x_n + y_n) \rightarrow \lambda$.
Let us suppose $\lambda \neq 0$.
Looking at the inequality :
$$f(x_n + y_n) \leq y_n f(x_n) + f(f(x_n))$$
You have a limit for the left term, but the right term cannot be minorated : contradiction.
Remark : If $\lambda = 0$ the argument not applies. So you have a problem if $a = b = 0$.
If you don't have continuity.
I recall some properties of the lim inf :
$$\lim \inf_{x \rightarrow - \infty} f(x+y) = \lim \inf_{x \rightarrow - \infty} f(x)$$
$$ \lim \inf_{x \rightarrow - \infty} a f(x) =  a \lim \inf_{x \rightarrow - \infty} f(x) \text{ if } a \geq 0 $$
$$ \lim \inf_{x \rightarrow - \infty} a f(x) =  a \lim \sup_{x \rightarrow - \infty} f(x) \text{ if } a \leq 0 $$
$$ \lim \inf_{x \rightarrow - \infty} f(x) + \lim \inf_{x \rightarrow - \infty} 
g(x) \leq \lim \inf_{x \rightarrow - \infty} f(x) + g(x) \leq \lim \inf_{x \rightarrow - \infty} f(x) + \lim \sup_{x \rightarrow - \infty} g(x)$$
Each inequality here might be strict.
Take the lim inf $x\rightarrow - \infty$ in the inequality $f(x+y) \leq y f(x) + f(f(x))$ to get :
$$a \leq ay + \lim \sup_{x \rightarrow - \infty} f(f(x)) \text{ for } y \geq 0$$
$$a \leq ay + \lim \inf_{x \rightarrow - \infty} f(f(x)) \text{ for } y \leq 0$$
So if you suppose $\lim \sup_{x \rightarrow - \infty} f(f(x)) < +\infty$ (which implies $\lim \inf_{x \rightarrow - \infty} f(f(x)) < +\infty$) you get a contradiction as soon as $a \neq 0$.
Take again the lim inf $x\rightarrow - \infty$ in the inequality $f(x+y) \leq y f(x) + f(f(x))$, but use this time $\lim \inf u(x) + v(x) \leq \lim \sup u(x) + \lim \inf v(x)$ to get :
$$a \leq by + \lim \inf_{x \rightarrow - \infty} f(f(x)) \text{ for } y \geq 0$$
$$a \leq by + \lim \sup_{x \rightarrow - \infty} f(f(x)) \text{ for } y \leq 0$$
With the same hypothesis $\lim \sup_{x \rightarrow - \infty} f(f(x)) < + \infty$, you get a contradiction as soon as $b \neq 0$.
It seems you need the assumption $\lim \sup_{x \rightarrow - \infty} f(f(x)) < + \infty$ to get something with your argument. ]]

For any $\lambda$ we can always choose a finite $y$ such that, where we get $\lambda\ > y\lambda+f(\lambda)$ which gives a contradiction. So, we deduce that  $\lim_{x\to -\infty}f(x)=-\infty$.

Ok, since the case $a = - \infty$, $b \neq - \infty$ can be covered by the preceding argument (you should have mentionned it).

Then, applying $x\to-\infty$ from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)\longrightarrow -\infty$. But, this contradicts with $f : \mathbb R → \mathbb R$.
So, we can deduce that $f(0)=0$.

For (X), you need to suppose $f$ continuous to make a rigorous proof (do it ! ). I must say your redaction looked messy because you did'nt skip lines. There is really little effort to do to improve this.
UPDATE :

Case 2. $f(0)<0$ (with the correct argument)
We have, from $f(x)\leq xf(0)+f(f(0))$ we deduce $\lim_{x\to +\infty}f(x)=-\infty.$ From $f(x + y) \leq yf(x) + f(f(x))$ we have:
$\begin{cases} f(x)\leq f(f(x)) \\ f(x) \leq xf(0)+ f(f(0)) \end{cases} \Longrightarrow f(x)\leq f(x)f(0)+f(f(0)) \Longrightarrow f(x)(1-f(0))\leq f(f(0))$.

Correct.

Then applying $x=f(0)$, we get $f(f(0))\leq 0$, which imply $f(x)\leq 0$, which gives $f(f(x))\leq 0$.

Nice.

In this case, we have $f(x)<0.$

It you be nice to add quantifiers. I think you mean : for all $x \in \mathbb{R}$.

Because, if $f(x)=0$, from $f(x)\leq f(f(x))$, we get $f(0)\geq 0$,

Be more precise : "if $f(x) = 0$ for some $x \in \mathbb{R}$". Ok for the argument.

which gives a contradiction. Applying $x\to-\infty$ from $f(0)\leq-xf(x)+f(f(x))$ we get $f(0)\longrightarrow -\infty$. Again a contradiction.

Are you supposing $-xf(x) \rightarrow - \infty$ ? It seems not to be necessarily the case (e.g. $f(x) = - \exp(-x)$) (XX)

So, we can deduce that $f(0)=0$.
Q.E.D.

You have to check (XX).
UPDATE 2 : (knowing $f < 0$) :

Then, from  $f(x + y) \leq yf(x) + f(f(x))$  we have:
$f(z)\leq(z-x)f(x)+f(f(x)) \Longrightarrow f(x) \leq (x-y)f(y)+f(f(y))\Longrightarrow 0\leq(f(y)-y)f(y) \Longrightarrow f(x)(f(x)-x)\geq 0 \Longrightarrow f(x) \leq x $

Great. This enables to conclude indeed. Good job.
