Definition [Uniformly Convex]
A Banach Space $(V, \Vert \cdot \Vert)$ is said uniformly convex:
For every $\epsilon >0$, exist $\delta > 0$ such that $\Vert x \Vert $ = 1 = $\Vert y \Vert $, and $\Vert x-y\Vert \ge \epsilon$ implies $$\left\Vert \frac {x+y}{2}\right\Vert \le 1-\delta.$$ Definition [Strictly Convex]
A Banach Space $(V, \Vert \cdot \Vert)$ is said strictly convex:
For every non-zero vector $x,y \in V$ linearly independent then $\Vert x+y \Vert < \Vert x \Vert + \Vert y \Vert$.
Theorem: If $X$ is finite dimensional Banach space, then $X$ strictly convex implies $X$ uniformly convex.
Proof: We will proof by using a contrapositive method.
Assume $X$ is not uniformly convex. Then exist $\epsilon_0 > 0$ for every $n \in \mathbb{N}$ and exist $x_n, y_n$ with $\Vert x_n\Vert = 1 =\Vert y_n\Vert$ such that $\Vert x_n - y_n\Vert \ge \epsilon_0$, but $$1 \ge \left \Vert \frac{x_n+y_n}{2} \right\Vert > 1- \frac{1}{n}.$$ Because $X$ is finite dimensional, then exist subsequence $(x_{nk}) \subset (x_n)$ and $(y_{nk}) \subset (y_n)$ with the limit of $(x_{nk}) \rightarrow x_0$ and $(y_{nk}) \subset (y_n) \rightarrow y_0$ fulfill $\Vert x_0 \Vert = 1 = \Vert y_0\Vert$when $k \rightarrow \infty$.
But $$\left \Vert \frac{x_0+y_0}{2} \right\Vert = \lim_{k \to \infty} \left \Vert\frac{x_{nk}+y_{nk}}{2} \right\Vert = 1.$$
As a result, $X$ is not strictly convex.
My Question:
Is my proof correct? How can we get the last equality?
You forgot the assumption of independence in the definition of strict convexity. What you can conclude at the end is that $x_0$ and $y_0$ are not independent. Hence, there is a scalar $c$ such that $x_0=cy_0$. But $x_0$ and $y_0$ are unit vectors so you get $|c|=1$ and $|1+c|=2$. The only complex number $c$ satisfying these two equations is $c=1$ so you get $x_0=y_0$. But $\|x_0-y_0\|\geq \epsilon$ so we have a contradiction.