First we prove the lower bound. Throughout the solution, we denote by $v_i$ and $l_i$ the number of vertices and edges of the $i$-th connected component, respectively. Each lattice point has 4 adjacent lattice points, and so the degree of each vertex in a connected component is at most 4. Clearly, the degree of the rightmost vertex in each row of a connected component is at most 3 and similar statement holds for the leftmost vertex in each row, and the topmost and the bottommost vertex in each column of a connected component. Thus, we have at least 4 vertices of degree 3 and so
$$2l_i = \text{Sum of Degrees in the } i\text{-th Connected Component} \le 4v_i - 4$$
Summing these inequalities for all connected components yields
$$2|A| \le 4 \sum v_i - 1 \Rightarrow \frac{l}{2} \le |V(A)| - |J(A)|$$
And this implies the left inequality.

Now we go to the right inequality. We know that there are $(n+1)^2$ vertices and $2n(n+1)$ edges in an $(n+1) \times (n+1)$ array of points. Let $n$ be the greatest positive integer such that $2n(n+1) \le l$. So we can write $l = 2n(n+1) + k$, where $0 \le k < 4(n+1)$ is an integer. We will introduce a graph with $l$ edges, satisfying the right inequality. Firstly, consider an $n \times n$ square, which has $2n(n+1)$ edges. If $k > 0$, we start to add some edges to the square. Suppose that the point $(0,0)$ is the leftmost point of the top side of the square. We add the edge connecting $(-1,0)$ to $(0,0)$. Then edges connecting $(-1,-1)$ to $(-1,0)$ and $(0,-1)$. In the same manner, if we need more edges we choose the next edge on the square perimeter (counterclockwise) and add two edges to the graph using that edge. Let $t$ be the number of sides in the square perimeter, which have been used in our process. Thus, we have added $t + 1 + \lfloor \frac{k-t}{2} \rfloor$ new vertices for constructing $k$ edges. We know $t \le \lfloor \frac{k}{2(n+1)} \rfloor$. Therefore, we have $(n+1)^2 + t + \frac{k-t}{2} + 1$ vertices in one connected component. So we must prove
$$(n+1)^2 + \frac{k+t}{2} < \frac{l}{2} + \sqrt{\frac{l}{2}} + 1 = n(n+1) + \frac{k}{2} + \sqrt{n(n+1) + \frac{k}{2}} + 1$$
Which is in turn equivalent to
$$n^2 + tn + \frac{t^2}{4} < n^2 + n + \frac{k}{2}$$
If $t = 1$, the inequality is obvious. And if $t = 2$, the inequality will be $n + 1 \le \frac{k}{2}$ which is again true because $2 = t = \lceil \frac{k}{2(n+1)} \rceil$ and so $k \ge 2(n + 1)$. So the assertion is proved.