Show that the interior of a convex set is convex

Question: Let $P\subseteq \mathbb{R}^n$ be a convex set. Show that $\text{int}(P)$ is a convex set.

I know that a point $x$ is said to be an interior point of the set $P$ if there is an open ball centered at $x$ that is contained entirely in $P$. The set of all interior points of $P$ is denoted by $\text{int}(P)$.

Also, to say that a set $P$ is convex means that if $x,y \in P$ then $tx+(1-t)y \in P$ for all $t \in (0,1)$.

How to go about the above proof?

Assume $U:=int(P)$ is not convex. There are $x_0,y_0\in U$ and $t_0\in (0,1)$ such that $z_0:=t_0x_0+(1-t_0)y_0\notin U$. Keep the segment $x_0y_0$ with fixed length and fixed at, say, $x_0$. Then $y$ is a continuous function of $z$, when you move $z$ and keep the segment $x_0y$ with the same length as $x_0y_0$. (These constrains are not really important, we are just fixing the ideas) Therefore, given a neighborhood $V\subset U$ of $y_0$ there is a neighborhood $W$ of $z_0$ such that for all points in $W$ the corresponding $y$ is in $V$. Since $z_0\notin int(P)$ there are points, say $z_1$, in $W$ that do not belong to $P$. Let $y_1$ be the corresponding $y$ for that $z_1$. Then $x_0$ and $y_1$ are two points of $P$ such that there is a point, $z_1$, in the segment $x_0y_1$ that is not in $P$. This is a contradiction because $P$ was convex. Therefore the non-interior point $z_0$ doesn't exist.

To prove the continuity of the correspondence above use triangle inequality.

I'll give a proof based on the following picture:

Suppose that $x$ and $y$ are interior points of a convex set $\Omega \subset \mathbb R^n$. Let $0 < \theta < 1$. We wish to show that the point $z = \theta x + (1 - \theta) y$ is in the interior of $\Omega$.

There exists an open ball $A$ centered at $x$ such that $A \subset \Omega$. Let $r$ be the radius of $A$, and let $B$ be the open ball of radius $\theta r$ centered at $z$.

Claim: $B \subset \Omega$.

Proof: Let $\hat z \in B$. There exists a point $\hat x \in \mathbb R^n$ such that $$ \hat z = \theta \hat x + (1 - \theta) y. $$ I will show that $\hat x \in A$. It will follow, by the convexity of $\Omega$, that $\hat z \in \Omega$. Since $\hat z$ is an arbitrary point in $B$, this will show that $B \subset \Omega$.

To complete the proof, note that \begin{align} \| \theta(\hat x - x)\| &= \| \theta \hat x - \theta x \| \\ &= \| \hat z - (1 - \theta) y - (z - (1 - \theta) y) \| \\ &= \| \hat z - z \| \\ &\leq \theta r. \end{align} It follows that $\| \hat x - x \| \leq r$, which shows that $\hat x \in A$.