Let $u_n=\frac {x_n} {\|x_n\|}$ and $v_n=\frac {y_n} {\|y_n\|}$. Let us show that $\|\frac {u_n+v_n} 2\| \to 1$. For this note that $\|x_n-u_n\|=\|x_n-\frac {x_n} {\|x_n\|} \|=|1-\|x_n\|| \to 0$. Similarly, $\|y_n-v_n\|\to 0$, Hence $|\|\frac {u_n+v_n} 2\| -\|\frac {x_n+y_n} 2\| |\leq \frac {\|x_n-u_n\|+\|y_n-v_n\|} 2 \to 0$. This proves that $\|\frac {u_n+v_n} 2\| \to 1$.
Now suppose $\|u_n-v_n\| \geq \epsilon$ for infinitely many $n$. Then $\delta(\epsilon) \leq 1-\|\frac {u_n+v_n} 2\| \to 0$ (through a subsequence), a contradiction. So we have proved that $\|u_n-v_n\| <\epsilon$ for $n$ sufficiently large . Since $\|x_n-u_n\| \to 0$ and $\|y_n-v_n\| \to 0$ we get $\|x_n-y_n\| \to 0$.
Luckily, I have found the proof which I post it below.
By normalizing, we can assume $M=1$. Assume the contrary that there exist $\varepsilon>0$ and a sequence $(x_n,y_n)$ with $|x_n|\le 1,|y_n|\le 1, |x_n-y_n|>\varepsilon$ such that
$$
0\le \frac{|x_n|^2+|y_n|^2}{2} -\left | \frac{x_n+y_n}{2} \right |^2 < \frac{1}{n}, \quad \forall n.
$$
Then
$$
\lim_n \left [\frac{|x_n|^2+|y_n|^2}{2} -\left | \frac{x_n+y_n}{2} \right |^2 \right ] =0.
$$
Notice that $({|x_n|}), ({|y_n|})$ are bounded. WLOG, we assume they are convergent, i.e., $|x_n| \to a$ and $|y_n| \to b$ with $a,b \ge 0$. Then
$$
\frac{a^2+b^2}{2} = \lim_n \left |\frac{x_n+y_n}{2} \right |^2 \le \lim_n \left [ \frac{|x_n|+|y_n|}{2} \right ]^2 = \left ( \frac{a+b}{2} \right )^2.
$$
This implies $a=b$ and thus
$$
\lim_n |x_n| = \lim_n |y_n| = \lim_n \left |\frac{x_n+y_n}{2} \right |=a.
$$
Let $\delta>0$ be the modulus of convexity of $\varepsilon$, i.e.,
$$
\left |\frac{x_n+y_n}{2} \right | \le 1-\delta, \quad \forall n.
$$
It follows that $a \le 1-\delta<1$. Next we pick $\eta,\lambda >0$ such that $\lambda(a+\eta) \le 1$ and $1-\delta < \lambda a <1$. WLOG, we assume $|x_n|,|y_n| \le a+\eta$ for all $n$. We define $x_n' := \lambda x_n$ and $y_n':=\lambda y_n$. It follows that $|x'_n|, |y'_n|\le 1$ and $|x_n'-y_n'| \ge \varepsilon$. We also get
$$
\lim_n \left |\frac{x'_n+y'_n}{2} \right | = \lambda a>1-\delta,
$$
which is a contradiction. This completes the proof.
Best Answer
The other direction is very tricky.
For suppose that the Banach space is not uniformly convex, then there are \begin{align*} \|x_{n}\|&\leq 1\\ \|y_{n}\|&\leq 1\\ \|x_{n}-y_{n}\|&\geq\epsilon_{0}\\ \left\|\dfrac{x_{n}+y_{n}}{2}\right\|&>1-\dfrac{1}{n}. \end{align*} Consider $u_{n}=x_{n}/\|x_{n}\|$ and $v_{n}=y_{n}/\|y_{n}\|$, then $\|u_{n}\|=\|v_{n}\|=1$ and \begin{align*} \left\|\dfrac{u_{n}+v_{n}}{2}\right\|\leq\dfrac{1}{2}(\|u_{n}\|+\|v_{n}\|)=1, \end{align*} and \begin{align*} \left\|\dfrac{u_{n}+v_{n}}{2}\right\|&\geq\left\|\dfrac{x_{n}+y_{n}}{2}\right\|-\left\|\dfrac{x_{n}}{2\|x_{n}\|}-\dfrac{x_{n}}{2}\right\|-\left\|\dfrac{y_{n}}{2\|y_{n}\|}-\dfrac{y_{n}}{2}\right\|\\ &>1-\dfrac{1}{n}-\dfrac{1}{2}\|x_{n}\|\left(\dfrac{1}{\|x_{n}\|}-1\right)-\dfrac{1}{2}\|y_{n}\|\left(\dfrac{1}{\|y_{n}\|}-1\right)\\ &=\dfrac{1}{2}(\|x_{n}\|+\|y_{n}\|)-\dfrac{1}{n}\\ &\geq\left\|\dfrac{x_{n}+y_{n}}{2}\right\|-\dfrac{1}{n}\\ &>1-\dfrac{2}{n}, \end{align*} so \begin{align*} \left\|\dfrac{u_{n}+v_{n}}{2}\right\|\rightarrow 1. \end{align*} By assumption, then $\|u_{n}-v_{n}\|\rightarrow 0$. But then \begin{align*} \epsilon_{0}&\leq\|x_{n}-y_{n}\|\\ &\leq\left\|x_{n}-\dfrac{x_{n}}{\|x_{n}\|}\right\|+\left\|y_{n}-\dfrac{y_{n}}{\|y_{n}\|}\right\|+\|u_{n}-v_{n}\|\\ &=2-\|x_{n}\|-\|y_{n}\|+\|u_{n}-v_{n}\|. \end{align*} Note that $\|x_{n}\|+\|y_{n}\|>2-2/n$, hence \begin{align*} \epsilon_{0}<\dfrac{2}{n}+\|u_{n}-v_{n}\|, \end{align*} by taking $n\rightarrow\infty$, we obtain a contradiction.