To find the area bounded by the lines x = 0, y = 0, and xy - 2 = 0, we can start by graphing these lines on a coordinate plane.

Graphing the lines:
First, let's graph the line x = 0. This is a vertical line passing through the y-axis at x = 0.
Next, let's graph the line y = 0. This is a horizontal line passing through the x-axis at y = 0.
Lastly, let's graph the equation xy - 2 = 0. To graph this equation, we can rewrite it as y = 2/x. We can plot a few points on the graph by choosing different values for x. For example, when x = 1, y = 2/1 = 2; when x = 2, y = 2/2 = 1; when x = -1, y = 2/(-1) = -2. These points give us an idea of the shape of the graph. As x approaches positive or negative infinity, y approaches zero. The graph of y = 2/x is a hyperbola that opens upwards in the first and third quadrants.

Finding the points of intersection:
To find the points where these lines intersect, we can set their equations equal to each other.
For x = 0 and y = 0, the point of intersection is (0, 0).
For y = 2/x and y = 0, we can set 2/x = 0 and solve for x. Since 2 divided by any non-zero number is not equal to zero, we can conclude that these lines do not intersect.

Determining the bounded region:
Since the lines x = 0 and y = 0 form the x and y axes respectively, the bounded region is the region below the hyperbola y = 2/x and between the y-axis and the x-axis.

Calculating the area:
To calculate the area of this bounded region, we can integrate the function y = 2/x with respect to x from x = 1 to x = 2.
The integral ∫(2/x)dx from 1 to 2 can be evaluated as ln(2) - ln(1) = ln(2).

Therefore, the area bounded by the lines x = 0, y = 0, and xy - 2 = 0 is ln(2) square units, which is approximately equal to 0.693.

Since the options provided in the question are in square units, the closest option is 2 square units, which is option C.