A set $E \subset \mathbb{R}^d$ is convex if whenever $x,y \in E$ then $\lambda x + (1- \lambda) y \in E$ where $\lambda \in (0,1)$
Show that $ B(x) = \{ y \in \mathbb{R}^d : |x-y| < \epsilon \}$ is convex set.
${\bf Attempt}$
Take two elements $a,b \in B (x)$ and so $|a-x| < \epsilon / (2 \lambda + 1 )$ and $|b-x| < \epsilon / (2 \lambda + 1 )$
Now,
$$ |\lambda a + (1-\lambda) b - x | = | \lambda ( a - b) + (b-x) | \leq \lambda |a-x+x-b| + |b-x| \leq \lambda (|a-x| + |b-x| ) + |x-b| < \dfrac{2 \lambda \epsilon}{2 \lambda + 1} + \dfrac{ \epsilon }{2 \lambda + 1} = \epsilon $$
So, B(x) is convex
Is this a correct solution?
You are assuming that $|a-x|$ and $|b-x|$ are smaller than what they are in general if $a,b\in B(x)$. There is no need for that; you must only assume that $a,b\in B(x)$, that is $|x-a|<\varepsilon, |b-x|<\varepsilon$. Under these assumptions only, you must take $\lambda\in(0,1)$ and show that $\lambda a+(1-\lambda)b\in B(x)$, i.e. you want to show that $$|\lambda a+(1-\lambda)b-x|<\varepsilon.$$ This is only a matter of calculations:
$$|\lambda a+(1-\lambda)b-x|=|\lambda a+(1-\lambda)b -(\lambda+1-\lambda)x|=$$ $$|\lambda a-\lambda x+ (1-\lambda )b-(1-\lambda)x|\leq|\lambda (a-x)|+|(1-\lambda)(b-x)|=$$ $$=\lambda\cdot|x-a|+(1-\lambda)\cdot|b-x|<\lambda\cdot\varepsilon+(1-\lambda)\cdot\varepsilon=\varepsilon. $$