Proving that closed (and open) balls are convex

Let $X$ be a normed linear space, $x\in X$ and $r>0$. Define the open and closed ball centered at $x$ as $$ B(x, r) = \{y \in X : \Vert x − y\Vert < r\} $$ $$ \overline{B}(x, r) = \{y \in X : \Vert x − y\Vert \leq r\}. $$ Then $B(x, r)$ and $\overline{B}(x, r)$ are convex.

I tried to prove this, but either my calculation is incorrect, or I am on the wrong path:

I aim to show for the closed ball $\overline{B}(x,r)$ (for open ball I assume the proof is similar). Suppose $y,z \in \overline{B}(x, r)$. Then $\Vert x − y\Vert \leq r$ and $\Vert x − z\Vert \leq r$. We must show that $t \in [0,1]$ implies $ty + (1-t)z \in \overline{B}(x,r)$. But $t \in [0,1]$ implies $$ \Vert ty + (1-t)z - x\Vert = \Vert t(y-z) + z - x\Vert \leq |t| \Vert y-z\Vert + \Vert z-x\Vert \leq |t|(\Vert y-x\Vert + \Vert x-z\Vert) + \Vert z-x\Vert < |t|(2r) + r = r(2|t| + 1), $$ which is not necessarily $\leq r$. We probably wanted to end up with $< |t|r$ or $\leq |t|r$ as our final inequality.

Thanks in advance.

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 $\square$