Interval sorting that avoids covering
If no two intervals end at the same point then you can simply sort them in increasing order of their right endpoint, and all of them will be visible.

If more than two intervals can end at the same point, then sort the intervals $[a,b]$ in increasing order of $b$ and break ties in decreasing order of $a$. 

Assume w.l.o.g., that there are no identical intervals (you can discard multiple copies of the same interval, since at most one copy will ever be visible). We will show that the above algorithm is optimal with an exchange argument.
Suppose that an optimal order $S$ of the intervals places two intervals in a different order than the one computed by the algorithm. Let $[a,b]$ and $[a',b']$ be the first two consecutive intervals in $S$ that are in the "wrong" order, that is:


$b&gt;b'$; or
$b=b'$ and $a&lt;a'$.


we will show that $[a,b]$ and $[a',b']$ can be swapped without decreasing the number of visible intervals.

If $[a,b]$ is not visible, then swapping $[a',b']$ with $[a,b]$ cannot decrease the number of visible intervals. Therefore we assume that $[a,b]$ is visible.
Let $p \in [a,b]$ be the largest point that is not covered by any the intervals preceding $[a,b]$.

If $p \not \in [a',b']$, then swapping $[a',b']$ with $[a,b]$ yields a solution $S'$ in which (i) $[a,b]$ is visible in $S'$, and (ii) if $[a',b']$ is visible in $S$ then it is visible in $S'$. Assume then that $p \in [a',b']$.

If $[a',b']$ is not visible then swapping $[a',b']$ with $[a,b]$ yields a solution in which $[a',b'] \ni p$ is visible (therefore the number of visible intervals cannot decrease). Assume then that $[a',b']$ is visible.

If $[a,b] \supset [a',b']$ then $[a',b']$ would not be visible, therefore we are in the first case ($b&gt;b'$). By the choice of $p$, all the points of the subinterval $(p,b]$ of $[a,b]$ are not visible. Therefore $S$ must contain an interval $[a'',b'']$ that precedes $[a,b]$ and is such that $b'' \ge b$ and $a'' &gt; p$.
This cannot happen since $[a'', b'']$ and $[a, b]$ is a pair of intervals in the wrong order which appears before the pair $[a,b]$ and $[a', b']$ in $S$.