Intuition. If $x$ is inside the interval $I$, then we can find a small closed interval $C$ around $x$ which is completely contained in $I$; because $x$ is the limit of the sequence, all except perhaps for finitely many terms of the sequence will be in $C$. Then just look at the missing terms; since there are only finitely many of them, we can enlarge the interval $C$ until it includes all the missing terms.
The idea works if $x$ is not inside the interval as well, though one has to be a bit careful then.
Proof. There is one degenerate case to consider; if singletons count as intervals, then we must consider the possibility that $I=\{p\}$; in that case, the limit point and all terms of the sequence must equal $p$, so the result holds with $J=I=\{p\}$. We therefore assume that $I$ has nonempty interior.
Consider first the case in which $x$ is in the interior of the interval $I$. Then there exists $\epsilon_1\gt 0$ such that $(x-\epsilon_1,x+\epsilon_1)\subseteq I$; if $\epsilon= \frac{1}{2}\epsilon_1$, then $[x-\epsilon,x+\epsilon]\subseteq I$.
Since the sequence converges to $x$, there exists $N\gt 0$ such that for all $n\geq N$, $x_n\in (x-\epsilon,x+\epsilon)$.
Let $b = \max \{ x+\epsilon, x_1, x_2,\ldots, x_N\}$. I claim first that $x_k\leq b$ for all $k$. Indeed, if $k\leq N$, then $x_k\leq \max\{x+\epsilon,x_1\ldots,x_N\}$, since $x_k$ is one of the elements of this finite set. And if $k\gt N$, then by the choice of $N$, $x_k\in (x-\epsilon,x+\epsilon)$, hence $x_k\lt x+\epsilon \leq b$.
I further claim that $b\in I$. Indeed, since $[x-\epsilon,x+\epsilon]\in I$, then $x+\epsilon\in I$. Since $(x_n)$ is a sequence in $I$, then $x_1,\ldots,x_N$ are each in $I$. Since this is a finite set of elements of $I$, the maximum is in $I$.
So $b\in I$, and $x_n\leq b$ for all $n\in\mathbb{N}$.
Now let $a = \min\{x-\epsilon, x_1,\ldots,x_N\}$. Then $x_k\geq a$ for all $k$: if $k\leq N$, because $x_k$ is in the set; and if $k\geq N$, because then $x_k\in (x-\epsilon,x+\epsilon)$. And $a\in I$, because each of $x_1,\ldots,x_N$ is in $I$, and $[x-\epsilon,x+\epsilon]\subseteq I$. So $a\in I$, and $a\leq x_n$ for all $n$. Also, clearly $a\lt b$, since $a\leq x-\epsilon\lt x+\epsilon\leq b$.
Since $I$ is an interval, $a\in I$, and $b\in I$, then $[a,b]\subseteq I$.
So we have the closed interval $J=[a,b]$, contained in $I$, and $a\leq x_n\leq b$ for all $n$, as desired.
If $x$ is not in the interior of $I$, then it is an endpoint of $I$. Suppose first that $x$ is the upper endpoint of $I$. Then there exists $\epsilon\gt 0$ such that $[x-\epsilon,x]\subseteq I$. Proceed then as above, working in the interval $(x-\epsilon,x+\epsilon)$, keeping in mind that all $x_i$ will necessarily lie in $(x-\epsilon,x]$. If $x$ is the lower endpoint of $I$ instead, then again we can find $\epsilon\gt 0$ such that $[x,x+\epsilon]\subseteq I$, and we can proceed the same way again.