If $a<b$, $c<d$ then $a+c<b+d$? This is how I managed to prove it: 

I know $a \lt b$ and $c \lt d$ thus, $b-a$ and $d-c$ are real positive numbers. Then, $b-a + d-c \gt 0$ and because of this $a+c \lt b+d$ 

Did I prove it right?
your way is correct,another one is 
add both side $c$ to the $a&lt;b$ so that $$a+c&lt;b+c\\ $$ 
then add $b$ to the $c&lt;d$ so that $$c+b&lt;d+b$$
finally we can conclude that $$a+c&lt;b+d$$
I would like to give a sweet and simple proof:


$a$ &lt; $b$ ---------------------(1)
$c$ &lt; $d$ ---------------------(2)
now simply add the two equations:
$a+c$ &lt; $b+d$
hope it helps you!