If the second $\delta(\varepsilon)$ is allowed to differ from the first one, then there is a simple implicit argument: Suppose the contrary, then there is a sequence $X_n$ of 2-dimensional normed spaces satisfying the definition with the same function $\delta(\varepsilon)$ and points $x_n,y_n\in X_n$ with $\|x_n\|\le 1$, $\|y_n\le 1\|$, $\|x_n-y_n\|\ge\varepsilon$ but $\|(x_n+y_n)/2\|\ge 1-\delta_n$ where $\delta_n\to 0$. Since the Banach--Mazur compactum is compact, there is a converging subsequence, and the limit space satisfies the definition for the same $\delta(\varepsilon)$ but contains two points $x,y$ with $\|x\|\le 1$, $\|y\|\le 1$, $\|x-y\|\ge\varepsilon$ and $\|(x+y)/2\|\ge 1$, a contradiction.

In fact, you can always choose the same $\delta(\varepsilon)$ in the second case provided that $\dim X\ge 2$. Suppose the contrary, then there are points $x,y\in X$ such that $\|x\|\le 1$, $\|y\|\le 1$, $\|x-y\|\ge\varepsilon$ but $\|(x+y)/2\|=1-\delta_1$ where $\delta_1<\delta=\delta(\varepsilon)$. We may assume that $X$ is 2-dimensional (otherwise restrict to a 2-dimensional subspace containing $x$ and $y$). Fix $\delta_1$ and from all such pairs $x,y$ choose one that minimize $\big|\|x\|-\|y\|\big|$. I claim that this minimizing pair satisfies $\|x\|=\|y\|$.

Suppose the contrary: let $\|x\|>\|y\|$. Denote $z=(x+y)/2$, $v=(x-y)/2$. If $v$ is proportional to $x$, choose any $v'$ with $\|v'\|=\|v\|$ such that $\|z+v\|\ne \|z\|\pm\|v\|$. Then the points $x'=z+v'$ and $y'=z-v'$ show that $x$ and $y$ did not minimize $\big|\|x\|-\|y\|\big|$. If $v$ is not proportional to $x$, choose a vector $w$ parallel to a supporting line to the unit sphere of $\|\cdot\|$ at the point $v/\|v\|$. Note that $w$ cannot be parallel to a supporting line at $x/\|x\|$, so either $\|x+tw\|<\|x\|$ or $\|x-tw\|<\|x\|$ for a sufficiently small $t>0$. Hence the points $x'=x+tw$ and $y'=y-tw$ or $x'=x-tw$ and $y'=y+tw$ provide a counter-example with $\big|\|x'\|-\|y'\|\big|<\big|\|x\|-\|y\|\big|$.

Thus the minimizing pair satisfies $\|x\|=\|y\|$. Multiplying by $\|x\|^{-1}$ we get a counter-example with $\|x\|=\|y\|=1$.