Show that $f$ is Lipschitz with Lipschitz constant equal to $1$ Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function s.t. for any $a\in \mathbb{R}$, there exists a strictly decreasing sequence $\{a_n\}$ s.t.$$\lim_{n\to\infty} a_n=a\ \text{and}\ |f(a_n)-f(a)|\leq a_n- a.$$
Show that $$|f(x)-f(y)|\leq|x-y|. \quad \forall x, y\in \mathbb{R}$$
My idea: 
I suppose $x<y$ and I consider $$x<x+  \frac{y-x}{n}< x+ \frac{2(y-x)}{n}< \cdots <y.$$
I don't know how I can continue this idea.
 A: Let $x<y$ be two real numbers.
We consider the set $L$ of all strictly increasing sequences $(l_i)$ such that 
$l_0=x$, for all $i$, $l_i\leq y$ and  $|f(l_{i+1})-f(l_{i})|\leq l_{i+1}-l_{i} $.
1) This set is not empty : from your condition, with $x=a$, it exists a sequence strictly decreasing $(a_i)$ etc., so, at least, with $l_0:=x$, and $i$ such that $a_i\leq y$, $l_1:=a_i$, we have 
$$|f(l_{1})-f(l_{0})|\leq |l_{1}-l_{0}|  $$
 We choose then $a=l_1$, and we construct $l_2$, with the same method, etc...
2) A bounded increasing sequence is convergent, so every sequence of $L$ has a limit $\leq y$.
We consider the set $\Lambda$ of the limits of all the sequence of $L$. Then we claim (since $f$ is continuous) that $\sup(\Lambda)=y$ and that gives $|f(x)-f(y)|\leq |y-x|$.
Indeed, if we suppose $\sup(\Lambda)=z<y$. The nature of the set $L$ implies that we have the following property : for all $\varepsilon>0$, there exists $u_n$ such that 
$$ z-\varepsilon <u_n \leq z, \quad \text{and} \quad |f(x)-f(u_n)|\leq |x-u_n|\leq z-x $$ 
So, we have a sequence $u_n\rightarrow z$, such that,
$$ |f(x)-f(u_n)| \leq z-x$$
so, by continuity of $f$,
$$ |f(x)-f(z)| \leq z-x \tag{1}$$
But here, we could construct a new sequence $(v_n)$ of $L$ with $l_0=x$ and $l_1=z$, $l_2>l_1$, etc... so $\sup(\Lambda)$ would be $>z$. Contradiction.
$\sup(\Lambda)=z= y$, so, we have   directly (it is just (1))
$$ |f(x)-f(y)| \leq y-x$$
