Well of course work is a scalar because that is how it is defined as a scalar product between the force and its displacement. But going a bit further if one wants to answer the question "why is work a scalar" one need to remember that basically work represent in its essence a change in energy, and thus the nature of work is related to the nature of energy itself.

`W =Delta(E)`

Now, why is energy a scalar? Motion of a mass has two quantities associated with it: linear momentum which is a vector and energy which is a scalar. Thus energy is the second physical quantity that shows how much motion a mass has. Since already we have a vector defined to show the quantity of motion (linear momentum) we need also another scalar quantity (energy) to describe the state of motion of a mass. Since linear momentum is always conserved in motion processes (like collision) we need energy (defined as a scalar) to show that some motion processes differ from other processes in some aspects (like for example inelastic collisions that differ from elastic collisions).

**This is why work is a scalar: its definition comes from energy considerations, and energy is a scalar.**

Observation:

Going a bit further with why energy is a scalar, we need to remember that the total energy is related to mass by the well known relation `E=m*c^2` and since the mass is always a scalar, and the speed is squared, the energy will be also a scalar.

You need to remember that work is defined as the scalar product between the force vector and the displacement vector, such that:

`W = bar F*bar d`

You need to remember the definition of the scalar product of two vectors, such that:

`bar u*bar v = |bar u|*|bar v|*cos theta`

`theta = (hat(bar u, bar v))`

You need to notice that the scalar product is a scalar quantity since is a product of magnitudes of the vectors `bar u` and `bar v` times the cosine of angle `theta` between the vectors.

**Hence, reasoning by analogy, since the work is the dot product between the force vector and the displacement vector, `W = |bar F|*|bar d|*cos theta` , thus, the work is a scalar quantity.**