Assume that the random variable $U$ follows the exponential distribution with parameter $\lambda$; also, suppose that random variable $X$ follows the general distribution $f_X(x)$. For $U>X>0$, i try to find distribution of $V=U-X$ as follows:
$$ \begin{split} H_V(z) &=P(U-X \leq z|U>X)=\frac{P(X<U\le X+z)}{P(U>X)}\\ P(X<U\le X+z) &= \int_0^\infty P(X<U<X+z|X=x)f_X(x)dx \\ &= \int_0^\infty P(x<U<x+z)f_X(x)dx \\ &= \int_0^\infty e^{-\lambda x }(1-e^{-\lambda z})f_X(x)dx \\ &= \left(1-e^{-\lambda z}\right) \int_0^\infty e^{-\lambda x }f_X(x)dx \end{split} $$ Also: $$ \begin{split} P(U>X) &= \int_0^\infty P(U>X|X=x)f_X(x)dx \\ &= \int_0^\infty P(U>x)f_X(x)dx \\ &= \int_0^\infty e^{-\lambda x}f_X(x)dx \end{split} $$
Therefore we have:
$$ H_V(z) = P(U-X \leq z|U>X) = \frac{(1-e^{-\lambda z})\int_0^\infty e^{-\lambda x }f_X(x)dx} {\int_0^\infty e^{-\lambda x}f_X(x)dx} = 1-e^{-\lambda z} $$
and
$$h_v(z)=\frac{dH_V(z)}{dz}$$
This is true if the variables are not dependent. But if the variables are dependent, this relationship is not correct. For example for $X=\frac{U}{2}$ then $V=U-X=\frac{U}{2}$, which follows exponential distribution with parameter $\frac{\lambda}{2}$.
Therefore, for dependent variables $U$ and $X$ can I say that $V=U-X$ still follows an exponential probability distribution but with a different parameter? if yes or not, how to prove it.
You statement is a little imprecise because when you write
it's not clear if that's a sure event, an event that the random variables verify a priori with probability one (that is $P(U> X)=P(X> 0)=1$) or not. (For example, the event $U>0$ is indeed a sure event).
What follows seems to imply that $U> X$ is not a sure event, only a condition. But what about the other event ($X>0$)?. If that's also a condition, you should write it down as such. Elsewhere, you should stipulate that $f_X$ is not totally general but restricted to positive values. Otherwise the limit of the first integral would be wrong.
Assuming the second alternative ($f_X$ is restricted to $\mathbb R_{>0}$), your result is correct (I don't know why you write $z$ instead of $v$ but that's not important) but also not surprising. If $U$ is exponential then it's memoryless , so thant $P(U>a+b|U>b)=P(U>a)$. And your result amounts to $$P(U> X+ z \mid U>X)=P(U>z)$$ which is an interesting generalization (already seen here), but basically the same thing.
Now, this is only true when $U,X$ are independent, because elsewhere the conditioning event would give some more (or less) information about $U$.
A simple counterexample. Let $X=\frac12 U$ (another exponential). Then the condition is a sure event, it can be dismissed and we get $$P(U> X+ z \mid U>X) =P(U> \frac12 U+ z)=P(U>2z) \ne P(U>z) $$