What are the minimum conditions for an open ball in $(X, d) $, $X$ is a linear space to be convex? In a normed linear space $(X,\Vert • \Vert  )$ any open ball $B(x_0, r) $ is convex.
My attempt:$B(x_0, r)=\{x\in X;\Vert x-x_0 \Vert<r\}$
$B(x_0, r)=x_0+rB(0, 1) $
So, it is enough to prove that open unit ball $B(0, 1) $ is convex.
For any two points $x, y\in B(0, 1)$ and $\lambda \in [0, 1]$,
$\Vert \lambda x +(1-\lambda) y\Vert$
$\le \lambda\Vert x\Vert+(1-\lambda)\Vert y\Vert$
$< \lambda •1 +(1-\lambda) •1= 1 $
Hence, open unit ball is convex.
Let's $(X, d) $ be a metric space where $X$ is a linear space.
My Question is -

*

*Any open ball in $X$ is convex. (T/F)

*What are the minimum conditions/assumptions to be insisted on $X$ for the open ball to be convex?

To prove 1) is false,
I have to find a metric on a linear space which is not induced by any norm on $X$.
I know in a linear space$X$ , a metric $d$ is induced by a norm if-
(I) $d(x+z, y+z) =d(x, y)$ and
(II) $d(\lambda x, \lambda y) =|\lambda | d(x, y) $ for any $x, y, z \in X$ , $\lambda$ any scalar.
So, I have to choose a metric on linear space $X$ which violates one of (I) or (II).
Please elaborate it with some example.
For, question 2), I have no idea. Please add some hints. Thanks.
 A: For your first question, the statement is indeed false. Let $X=\mathbb{R}^2$ be equipped with the metric (check that this is a metric: use that $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$)
$$d(x,y)=d((x_1,x_2),(y_1,y_2))=\sqrt{|x_1-y_1|}+\sqrt{|x_2-y_2|}. \tag{1}$$
Then I claim that $B(0,1)$ is not convex (in fact, you can generalise this to $B(x,r)$ for any $x\in X$ and $r>0$ but this is not necessary). Namely, take $y=(3/4,0)$ and $z=(0,3/4)$. Then $y,z\in B(0,1)$ (check this) but $(1/2)y+(1/2)z=(3/8,3/8)\not\in B(0,1)$ since
$$d((0,0),(3/8,3/8))=\sqrt{3/8}+\sqrt{3/8}=\frac{\sqrt{6}}{2}>1.$$

For your second question, the answer is that if $X$ is a linear space and $d$ is a metric on $X$, then open balls are convex if and only if
$$d(x,ty+(1-t)z)\leq \max\{d(x,y),d(x,z)\} \tag{2}$$
for all $x,y,z\in X$ and $t\in [0,1]$.
It is easy to verify that this inequality implies that open balls $B(x,r)$ are convex for all $x\in X$ and $r>0$ (I'll leave this as an exercise, as it's just a matter of understanding definitions).
Conversely, suppose there exists $x,y,z\in X$ and $t\in [0,1]$ such that
$$d(x,ty+(1-t)z)>\max\{d(x,y),d(x,z)\}.$$
In this case we may choose an $r$ such that
$$\max\{d(x,y),d(x,z)\}<r<d(x,ty+(1-t)z).$$
However, this implies that $y,z\in B(x,r)$ but $ty+(1-t)z\not\in B(x,r)$. Hence $B(x,r)$ is not convex for this $r$.

You can of course verify that the NLS case (i.e. $d(x,y)=\|x-y\|$) satisfies the inequality $(2)$, and that the metric in $(1)$ does not satisfy $(2)$.
