Substituting $x,y:=0$ gives $f(0)=0$. Letting $y:=-x$, we conclude that $f$ is odd. Now, $y:=0$ makes $f(x) \leq x$. However, that would mean $-f(x)=f(-x) \leq -x$, or $x \leq f(x)$. Thus, only the identity satisfies the inequalities.

WakeUp wrote: Determine all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying $f(x+y)\le f(x)+f(y)\le x+y$ for all $x,y\in\mathbb{R}$. Let $P(x,y)$ the assertion of the given Functional Inequality. $P(0,0)$ $$f(0) \le 2f(0) \le 0 \implies f(0)=0$$$P(x,-x)$ $$0 \le f(x)+f(-x) \le 0 \implies f(x)=-f(-x) \implies f \; \text{odd}$$$P(x,0)$ $$f(x) \le f(x) \le x \implies f(x) \le x$$$P(-x,0)$ $$f(-x) \le -x \implies -f(x) \le -x \implies x \le f(x)$$Hence the only solution is: $\boxed{f(x)=x \; \forall x \in \mathbb R}$ Thus we are done

$P(0,0)\Rightarrow f(0)\le2f(0)\le0\Rightarrow f(0)=0$ $P(x,-x)\Rightarrow 0\le f(x)+f(-x)\le0\Rightarrow f(x)+f(-x)=0$ $P(x,0)\wedge P(-x,0)\Rightarrow x\le f(x)\le x\Rightarrow\boxed{f(x)=x}$, which works.