I have the inequalities:
\begin{align} b>c\\ a<b \end{align} multiplying by $-1$ the first and then substracting from the second:
$$-b<-c$$ $$ a< b$$ $$-b-a<-c-b$$
that would imply $-a<-c$ or $a>c$.
But if I take $b=3, c=2$ and $a=1$, it doesn't work.
On
If you subtract you must subtract the larger from the smaller and the smaller from the larger.
If $a <b $ and $c <d $ you can conclude $a-d<b-c $ but you can't conclude anything about $a-c $ or $b-d$.
After all $10 <11$ and $2 <8$ but $10-2\not < 11-8$
Addition works because if $a <b $ and $c <d$ the $a+c <b+c <b+d$. you can do that because both $a,c $ are less than $b,d$.
But if we try to do $a+d<b+c $ that fails utterly. $a <b $ but $d>c $ so adding them could do anything.
Subtraction is different because we are taking away. If $m <n $ then $k-m>k-n $. This makes sense because taking away something small will leave something big.
So if we have $a <b $ and $c-d $ then $a-d <a-c <b-c $.
But we can't say anything about $a-c $ and $b-d $ because the values being compared are in consistant..
Think about how the numbers $a$, $b$, $c$ are located on the real number line.
The inequality $b>c$ says that $b$ is to the right of $c$, which means that $c$ is to the left of $b$.
The inequality $a<b$ says that $a$ is to the left of $b$.
So, all we know is that both $a$ and $c$ are to the left of $b$, and this tells us nothing about the relationship between $a$ and $c$.
As the comments indicate, the flaw in your reasoning is the subtraction step.