Every principal ideal domain satisfies ACCP.

Every principal ideal domain $D$ satisfies ACCP (ascending chain condition on principal ideals)

Proof. Let $(a_1) ⊆ (a_2) ⊆ (a_3) ⊆ · · ·$ be a chain of principal ideals in $D$. It can be easily verified that $I = \displaystyle{∪_{i∈N} (a_i)}$ is an ideal of $D$. Since $D$ is a PID, there exists an element $a ∈ D$ such that $ I = (a)$. Hence, $a ∈ (a_n)$ for some positive integer $n$. Then $I ⊆ (a_n) ⊆ I$. Therefore, $I = a_n$. For $t ≥ n$, $(a_t) ⊆ I = (a_n) ⊆ (a_t)$. Thus, $(a_n) = (a_t)$ for all $t ≥ n$.

I have prove $I$ is an ideal in the following way:-

Let $ x,y\in I$. Then there exist $i,j \in \mathbb{N}$ s.t. $x \in (a_i)$ & $y \in (a_j)$.
Let $k \in \mathbb{N}$ s.t $k>i,j$.
Then $x \in (a_k)$ & $y \in (a_k)$.
as $(a_k)$ is an ideal $x-y \in (a_k)\subset I$ and $rx,xr \in (a_k)\subset I$.
So $I$ is an ideal.

Is it correct?

Your proof is right but you can let t = max(i,j) and any k > t.

Yes. Said more simply $\ (a_1) \subseteq (a_2) \subseteq \cdots\subseteq (a_1,a_2,a_3,...)\stackrel{\rm PID} = (c_1 a_1 +\cdots + c_k a_k) \subseteq (a_k)$