Let [x] denote the greatest integer function. Then match List-I with List-II:

The correct option is: (A)- (II), (B)- (I), (C)- (III), (D)- (IV)

Let us analyze each function in List-I one by one and match it with the correct property in List-II based on the visual details and formulas provided in the lists:

(A) Analysis of $f\left(x\right)=|x-1|+|x-2|$:
The absolute value function $\left|x\right|$ is continuous everywhere on the real number line.
Since the sum of two continuous functions is also continuous, the function $f\left(x\right)=|x-1|+|x-2|$ is continuous everywhere.
Thus, (A) matches with (II).

(B) Analysis of $g\left(x\right)=x-\left|x\right|$:
We can rewrite this piecewise function as:
For $x\ge 0$:
$$g\left(x\right)=x-x=0$$
For $x<0$:
$$g\left(x\right)=x-(-x)=2x$$
Now let us check the differentiability at $x=0$:
Right-hand derivative (RHD) at $x=0$:
$$\underset{h\to 0^{+}}{lim}\frac{g(0+h)-g\left(0\right)}{h}=\underset{h\to 0^{+}}{lim}\frac{0-0}{h}=0$$
Left-hand derivative (LHD) at $x=0$:
$$\underset{h\to 0^{-}}{lim}\frac{g(0+h)-g\left(0\right)}{h}=\underset{h\to 0^{-}}{lim}\frac{2h-0}{h}=2$$
Since $\text{RHD}\ne \text{LHD}$ at $x=0$, the function is not differentiable at $x=0$. For all other real values of $x$, the function is linear and therefore differentiable.
Thus, $g\left(x\right)$ is differentiable everywhere except at $x=0$, which means (B) matches with (I).

(C) Analysis of $h\left(x\right)=x-\left[x\right]$:
Here $\left[x\right]$ represents the greatest integer function. The function $h\left(x\right)=x-\left[x\right]$ is the fractional part function, denoted by $\left\{x\right\}$.
This function has a discontinuity at every integer. Let us verify its continuity at $x=1$:
$$\underset{x\to 1^{-}}{lim}(x-[x\left]\right)=1-0=1$$
$$\underset{x\to 1^{+}}{lim}(x-[x\left]\right)=1-1=0$$
Since the left-hand limit and right-hand limit at $x=1$ are not equal, $h\left(x\right)$ is discontinuous at $x=1$.
Any function that is discontinuous at a point is also non-differentiable at that point. Thus, $h\left(x\right)$ is not differentiable at $x=1$.
Thus, (C) matches with (III).

(D) Analysis of $p\left(x\right)=x\left|x\right|$:
We can rewrite this function as:
For $x\ge 0$:
$$p\left(x\right)=x^2$$
For $x<0$:
$$p\left(x\right)=-x^2$$
In a neighborhood around $x=1$ (where $x>0$), the function is represented by $p\left(x\right)=x^2$.
Since $x^2$ is a polynomial function, it is differentiable everywhere on its domain, and specifically at $x=1$ where its derivative is $p^{\prime }\left(1\right)=2\left(1\right)=2$.
Thus, $p\left(x\right)$ is differentiable at $x=1$, which means (D) matches with (IV).

Conclusion:
Combining the matches:
• (A) matches with (II)
• (B) matches with (I)
• (C) matches with (III)
• (D) matches with (IV)