Prove there exists $M<1$ such that $|f(x)-f(0)|\le M|x| $ holding for every $ x \in \mathbb{R}$. 
Suppose $f(x)$ be bounded and differentiable over $\mathbb R$, and
$|f'(x)|<1$ for any $x$. Prove there exists $M<1$ such that
$|f(x)-f(0)|\le M|x|$ holding for every $x \in \mathbb{R}$.

Probably, we may consider applying MVT, say $$\left|\frac{f(x)-f(0)}{x-0}\right|=|f'(\xi)|<1,$$ but which can only imply $M\le 1$, not $M<1$ we wanted. How to improve the inequality?
 A: You can assume without loss of generality that $f(0) = 0$.
Suppose to the contrary that no such $M$ exists. Then for each $k \ge 2$ there exists a point $x_k$ satisfying $$|f(x_k)| > \left( 1 - \frac 1k \right) |x_k|.$$
Since $f$ is bounded there is a constant $C$ with $|f(x_k)| \le C$ for all $k$, implying that $|x_k| \le \dfrac{kC}{k-1} \le 2C$ for all $k$. Thus $\{x_k\}$ is bounded and has a convergent subsequence $\{x_{k_j}\}$. Denote the limit of this subsequence by $x$.  Since $f(x_{k_j}) \to f(x)$ and $$|f(x_{k_j})| > \left( 1 - \frac 1{k_j} \right) |x_{k_j}|$$  you may let $j \to \infty$ to discover that $|f(x)| \ge |x|$. According to the mean-value theorem there must be a point $\xi$ in between $0$ and $x$ with $|f'(\xi)| \ge 1$, contrary to hypothesis.
A: The function $g: \Bbb R \to \Bbb R$, defined as
$$
 g(x) = \begin{cases}
 \frac{f(x)-f(0)}{x-0} & \text{ if } x \ne 0  \\
f'(0) & \text{ if } x = 0  
\end{cases}
$$
is continuous, with $\lim_{x \to - \infty} g(x) = \lim_{x \to + \infty} g(x)= 0$.
It follows that $|g|$ attains its maximum at some point $x_0$, so that
$$
 |g(x)| \le |g(x_0)| = \left|\frac{f(x_0)-f(0)}{x_0-0}\right| = |f'(\xi_0)| =: M < 1
$$
for some fixed $\xi_0$, and all $x \in \Bbb R$.
