$\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$To complement YCor's answer, here is a counterexample using a $G$ which is neither linear nor simply connected. (YCor showed that, if $G$ is either linear or simply connected, the conclusion is true.) It is closely inspired by Tom Goodwillie's answer that didn't work.
Define $$H = \begin{bmatrix} 1&\RR&\RR/\ZZ \\ 0&1&\RR \\ 0&0&1 \end{bmatrix}$$ $$H_1 = \begin{bmatrix} 1&\RR&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \qquad H_2 = \begin{bmatrix} 1&0&0 \\ 0&1&\RR \\ 0&0&1 \end{bmatrix} \qquad Z_H = \begin{bmatrix} 1&0&\RR/\ZZ \\ 0&1&0 \\ 0&0&1 \end{bmatrix}.$$ Then it is easy to check that $(H_1, H_2) = Z_H$, more precisely, $$\left( \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}, \begin{bmatrix} 1&0&0 \\ 0&1&y \\ 0&0&1 \end{bmatrix}\right) = \begin{bmatrix} 1&0&xy \\ 0&1&0 \\ 0&0&1 \end{bmatrix}.$$
Now, let $G = H \times H$, and put $$G_1 = \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \subset H_1 \times H_1$$$$G_1 = \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix} 1&x&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \subset H_1 \times H_1 \subset G$$ $$G_2 = \begin{bmatrix} 1&0&0 \\ 0&1&y \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix} 1&0&0 \\ 0&1&\sqrt{2} y \\ 0&0&1 \end{bmatrix} \subset H_2 \times H_2.$$$$G_2 = \begin{bmatrix} 1&0&0 \\ 0&1&y \\ 0&0&1 \end{bmatrix} \times \begin{bmatrix} 1&0&0 \\ 0&1&\sqrt{2} y \\ 0&0&1 \end{bmatrix} \subset H_2 \times H_2 \subset G.$$ Then $G_1$ and $G_2$ are both closed in $G$, but $(G_1, G_2)$ is the immersed subgroup $(z, \sqrt{2} z)$ in $Z_H \times Z_H \cong (\RR/\ZZ)^2$, which is not closed.