Inequality with two absolute values
How to solve $|x-3|-|x-4|<x$? Let $f(x)=|x-3|-|x-4|$
I would begin by noting $|x-3|=x-3$ for $x \geq 3$ whereas $|x-3|=3-x$ for $x \leq 3$. Likewise, $|x-4|=x-4$ for $x \geq 4$ whereas $|x-4|=4-x$ for $x \leq 4$
if $x \geq 4$ then $x > 3$ hence $f(x)=x-3-(x-4)=1$ and we find $1<x$ which is true in this case. This puts $[4,\infty)$ in the solution set of the inequality.
if $x \leq 3$ then $x < 4$ hence $f(x)=3-x-(4-x)=-1$ and we face $-1<x$ which is true for each $x$ with $-1<x\leq 3$. This shows $(-1,3]$ is also in the solution set of the inequality.
if $3<x<4$ then $f(x) = x-3-(4-x)=2x-7$. Hence in this context we must solve $2x-7<x$ which gives $x<7$ which is true for each $x$ in the interval considered. This places $(3,4)$ in the solution set of the inequality.
Put all of this together, we obtain $(-1,3] \cup (3,4) \cup [4,\infty) = (-1, \infty)$ which is your answer. Usually I solve this sort of thing with a sign-chart method.
The standard semi-mechanical way to eliminate the absolute value signs is to divide the number line into segments. The critical point for $|x-4|$ is at $x=4$, and the critical point for $|x-3|$ is at $x=3$.
Suppose first that $x \ge 4$. Then $|x-4|=x-4$ and $|x-3|=x-3$. So we are looking at the inequality $(x-3)-(x-4)\lt x$, that is, at $1\lt x$, which is certainly true for $x\ge 4$.
Now suppose that $3\le x\lt 4$. Then $|x-4|=4-x$ and $|x=3|=x-3$, so we are looking at the inequality $(x-3)-(4-x)\lt x$, that is, $2x-7\lt x$. This simplifies to $x\lt 7$, which is certainly true in the interval $[3,4)$.
It is probably at this point that your calculation went astray. We were looking at the interval $[3,4)$ and asking which points in this interval satisfied our inequality. The manipulation told us that it was all points in this interval that satisfied $x\lt 7$. Well, they all do!
Finally, suppose that $x\lt 3$. Then $|x-4|=4-x$ and $|x-3|=3-x$. So we are looking at the inequality $(3-x)-(4-x)\lt x$. Calculate. The left side is $-1$, so for in the interval $(-\infty,3)$, the inequality holds precisely when $-1\lt x$.
Putting things together, we conclude that the original inequality holds (i) if $x\ge 4$; (ii) if $3\le x\lt 4$; and (ii) if $x\lt 3$ but $-1\lt x$. This complicated set of conditions can be summarized much more simply as $x\gt -1$.
There are other ways to describe the set of solutions. For example we could say that the set is $(-1,\infty)$.
What you have is almost correct, the last step is to restrict your solution to the corresponding region.
For example, for $x>1$, the answer must be in the $x>4$ region, so your answer for this region is $x>4$.
For $x>-1$, your answer must be in $x<3$ region, so your answer for this region is $-1<x<3$.
And for the last one, $x<7$, your answer must be in the corresponding region we had first, namely $3<x<4$, so your answer for this region is $3<x<4$.
Now draw these three answers on the number line and you'll have $x>-1$, the desired final answer.