$\Rightarrow \text{:}$ Assume $J$ is not an interval.
Let $x,y\in J$ and $z\in ℝ$ with $$x<z<y,x,y\in J\text{and}z\notin J\text{.}$$ Let $A=(-\infty ,z)\cap J$ and $B=(z,\infty )\cap J\text{.}$
Then $A$ and $B$ are open subsets of $J$ and $$A\ne \varnothing ,B\ne \varnothing ,A\cap B=\varnothing \text{and}A\cup B=J\text{.}$$ So $J$ is not connected.

$\Leftarrow \text{:}$ Assume $J$ is an interval.
To show: $J$ is connected.
Proof by contradiction.
Assume $J$ is not connected.
Let $A\subseteq J$ and $B\subseteq J$ be open subsets of $J$ such that $$A\cap B=\varnothing ,A\ne \varnothing ,B\ne \varnothing \text{and}A\cup B=J\text{.}$$ Then $f:J\to \{0,1\}$ given by $$f\left(z\right)=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}z\in A,\\ 1,& \text{if}\hspace{0.17em}z\in B\end{array}$$ is a continuous surjective function.
Let $x_1,y_1\in J$ with $f\left(x_1\right)=0$ and $f\left(y_1\right)=1\text{.}$
Switching $A$ and $B$ if necessary we may assume that $x_1<y_1\text{.}$
Construct sequences $x_1,x_2,\dots$ and $y_1,y_2,\dots$ by $$\begin{array}{c}{\displaystyle x_{i+1}=\frac{x_i+y_i}{2}\text{and}{y}_{i+1}=y_i,\text{if}f\left(\frac{x_i+y_i}{2}\right)=0,}\\ {\displaystyle x_{i+1}=x_i\text{and}{y}_{i+1}=\frac{x_i+y_i}{2},\text{if}f\left(\frac{x_i+y_i}{2}\right)=1\text{.}}\end{array}$$ By induction, $x_i\in J$ and $y_i\in J,$ and, since $J$ is an interval, $\frac{x_i+y_i}{2}\in J$ so that $f\left(\frac{x_i+y_i}{2}\right)$ is defined and $$x_{i+1}\in J\text{and}{y}_{i+1}\in J\text{.}$$ Also, $$f\left(x_{i+1}\right)=0,f\left(y_{i+1}\right)=1,x_i\le x_{i+1}<y_{i+1}\le y_i\text{and}|x_{i+1}-y_{i+1}|\le \frac{1}{2}|x_i-y_i|$$ so that $$|x_{i+1}-y_{i+1}|\le \frac{1}{2^i}|x_1-y_1|\text{.}$$ Since $ℝ$ is complete and the sequence $x_1,x_2,\dots$ is increasing and bounded by $y_1,$ $\underset{n\to \infty }{\text{lim}}{x}_n$ exists in $ℝ\text{.}$
Since $ℝ$ is complete and the sequence $y_1,y_2,\dots$ is decreasing and bounded by $x_1,$ $\underset{n\to \infty }{\text{lim}}{y}_n$ exists in $ℝ\text{.}$
Since $\underset{n\to \infty }{\text{lim}}|x_n-y_n|=0$ then $\underset{n\to \infty }{\text{lim}}{x}_n=\underset{n\to \infty }{\text{lim}}{y}_n\text{.}$
Let $$z=\underset{n\to \infty }{\text{lim}}{x}_n=\underset{n\to \infty }{\text{lim}}{y}_n\text{.}$$ Since $x_1\le x_2\le \cdots \le x_n<y_n\le y_{n-1}\le \cdots \le y_1$ for $n\in {\mathbb{Z}}_{>0}$ then $$x_1<z<y_1\text{.}$$ Since $J$ is an interval, $z\in J\text{.}$
Since $f$ is continuous, $$0=\underset{n\to \infty }{\text{lim}}f\left(x_n\right)=f\left(z\right)=\underset{n\to \infty }{\text{lim}}f\left(y_n\right)=1\text{.}$$ This is a contradiction.
So $J$ is connected.

$\square$