Prove $aGiven: $a<b,c<d$
To Prove: then $|a-c|+|b-d|<|a-d|+|b-c|$
Proof(Attempt):
$$|a-c|+|b-d|=|a-c+d-d|+|b-c+c-d|\\\le |(a-d)+(c-d)|+|(b-c)+(c-d)|\\
\stackrel{?}\le|a-d| +|b-c|+\underbrace{(c-d)+(d-c)}_{?} $$
 A: We can assume $a=0$ (otherwise reduce all numbers for $a$)
$$|c|+|b-d|<|d|+|b-c| \;\;\;/ ^2$$
$$c^2+b^2+d^2-2bd+2|c||b-d| < d^2+b^2+c^2-2bc +2|d||b-c|$$
$$-bd+|c||b-d| < -bc +|d||b-c|$$
$$|c||b-d|-|d||b-c| < b(d-c)$$
Last one is true since we have, by triangle inequality ($|x|-|y|\leq|x-y|$): 
$$|c||b-d|-|d||b-c| \leq |bc-dc -db+cd| = b|c-d| $$
A: You can only show that
$$ \tag{*}
|a-c|+|b-d| \le |a-d|+|b-c|
$$
and equality can hold, e.g. for $(a, b, c, d) = (1, 2, 3, 4)$.
More precisely, we have

Let $a < b$, $c < d$ be real numbers, and denote by $I, J$ the open intervals $I = (a, b)$ and $J = (c, d)$. Then
  $$ \tag{**}
 |a-c|+|b-d| + 2 m(I \cap J) = |a-d|+|b-c| 
$$
  where $m(I \cap J)$ is the length of the intersection of the intervals,
  or zero if intervals are disjoint.

It follows that $(*)$ holds, with equality exactly if the intervals $I, J$ are disjoint.

For the proof of $(**)$ consider two cases where $I$ and $J$
are disjoint, and a third case where they do intersect:
Case 1: $b \le c$. Then $m(I \cap J) = 0$ and
$$
|a-c|+|b-d| = (c-a) + (d-b) = (d-a) + (c-b) = |a-d|+|b-c| \, .
$$
Case 2: $d \le a$. Again $m(I \cap J) = 0$, and
$$
|a-c|+|b-d| = (a-c) + (b-d) = (a-d) + (b-c) = |a-d|+|b-c| \, .
$$
Case 3:  $b > c$ and $d > a$. Then $I \cap J$ is a non-empty
open interval with length
$$
 m(I \cap J) = \min(b,d) - \max(a, c) > 0 
$$
and
$$
|a-c|+|b-d| + 2m(I \cap J) \\
= \max(a, c) - \min(a, c) + \max(b, d) - \min(b,d) + 2\bigl(\min(b,d) - \max(a, c)\bigr) \\
= \max(b, d) + \min(b, d) - \max(a, c) - \min(a, c) \\
= (b + d) - (a + c) = (d-a) + (b-c) \\
= |a-d|+|b-c| \, .
$$
