Proving that closed (and open) balls are convex

Hint: $$ ty+(1-t)z-x=t(y-x)+(1-t)(z-x) $$

Suppose $x, y$ are in a unit closed ball. Consider $$||z||=||(1-a)x+ay||\leq (1-a)||x||+a||y||\leq 1-a+a=1.$$ So $||z||\leq 1$, so $z$ is in unit close ball.

Let $ B(a,r)=\{y \in X: \|y-a\| < r \}$ be an open ball in a Banach space $X$. Let $x,y \in B(x,r)$ then $\| x-a\| < r $ and $\| y-a\| < r $. $$\|x\|-\|a\| <\| x-a\| < r ~~ and~~ \|y\|-\|a\| <\| y-a\| < r$$ which implies \begin{equation} \|x\| < r + \|a\|,~ \|y\| < r + \|a\|~~~~~~ (1) \end{equation} for any $t \in [0,1]$; $x,y \in X$ using $(1)$ we have $$ \|tx+(1-t)y\| \leq t\|x\|+(1-t)\|y < t(r+\|a\|)+(1-t)(r+\|a\|)=r+\|a\|\\ $$ Since $a$ is the centre (Origin) of the ball $\|a\|=0$. Hence, the result $\square$