Sometimes, unfortunately, the method you are "supposed" to use is gross overkill. But I will write out a linear inequalities approach, under the assumption that the
number of fish eaten on any day is a non-negative integer.
Also, when your text said less than and your formula said $\le\;$, I assume it should be $\lt$. Similarly, if your text says more and your formula said $\ge\;$, I assume it should be $\gt$. So when there is conflict, I give priority to the words.
We have three variables, but one can be eliminated, using say $z=12-(x+y)$.
Since $x<y$, we have
(1) $x+1 \le y$.
Since $y \lt z$, we have $y+1 \le z$, and therefore $y+1 \le 12-(x+y)$. This can be rewritten as
(2) $x+2y \le 11$.
We have $z \lt x+y$, so $z+1 \le x+y$, and therefore $12-(x+y)+1 \le x+y$, giving
(3) $13 \le 2x+2y$.
Finally, we need to remember that $x$, $y$, and $12-(x+y)$ are all $\ge 0$.
That gives the inequalities
(4) $x \ge 0$, $y \ge 0$, $12 \ge x+y$.
Now we can draw lines as usual.
Much too much work, since it is clear that $x=3$, $y=4$, $z=5$ is the only solution! This is because if $x\lt y \lt z$ and $x+y+z=12$, we must have $z\gt 12/3$, so $z \ge 5$. If $z\ge 6$, then $x+y \le 6$, but $x+y \le z$ is not allowed. So $z=5$. Now we have $7$ fish, to be distributed among the two days. But the second day must be $\le 4$. To get a total of $7$ we need $y=4$, $x=3$.
Note: If everywhere we use your formula inequalities, not your text inequalities, all the $+1$'s need to be removed. That changes the inequalities, and we get a few more solutions. Which ones are OK depends on interpretation of the wording. The solutions could include some or all of $(0,6,6)$, $(1,5,6)$, $(2,4,6)$, $(2,5,5)$, $(3,3,6)$, $(3,4,5)$, and $(4,4,4)$.