It is not so hard, but it won't be able to generate enough high energetic particles.
The best example for a particle accelerator is a CRT (cathode ray tube), which you can find in every CRT monitor or TV. It can generate around $40\rm\,keV$ electrons. (LHC generates $3.5\rm\,TeV$ protons, thus it is around a hundred million times stronger).
Only a particle accelerator isn't enough, if you want to make experiments with it, you need some analytic/measurements devices also. In the current accelerators, they are nearly so complex and costly as the main accelerating device.
There is also a device capable to be built in home, it is the Farnsworth fusor:
Maybe it is not a particle accelerator in the classical sense, it creates enough strong field to be able to fuse deuterons (although it is doing this with terrible efficiency, around $10^{-8}$). You can see a Farnsworth fusor scematic below:
(source: fusor.net)
There is a whole community of home fusors which can be found here.
As you point out in your ray diagram, the point source no longer looks like a point source after the diffraction grating. You could get around this by collimating the input beam with another lens (which you did), with an aperture, or by moving the point source far away from the grating (effectively using the grating as an aperture). So, I think the answer to your question is no unless you are willing to better collimate the input source.
Two Diffraction Gratings
You could, however, use a second diffraction grating to reproduce the original image. To see this, consider one of the rays in your diagram with initial angle at the grating of $\theta_1$. After the grating the ray in the first order diffraction will have angle
$$
\theta_2=\arcsin\left(\frac{\lambda}{d}-\sin(\theta_1)\right),
$$
where $\lambda$ is the wavelength and $d$ is the grating's spacing. After the second grating the first order diffraction will have angle
$$
\theta_3=\arcsin\left(\frac{\lambda}{d}-\sin\left(\theta_2\right)\right)=\theta_1.
$$
So, the angle of the ray is restored to its original value after the second diffraction grating. I.E. the image of the point source is restored.
The position of the ray's which form an image will be shifted by an amount determined by the spacing between the two gratings and the wavelength. To see this consider the position of the ray along the optical axis of the system. For this ray $\theta_1=0$ and
$$
\theta_2=\arcsin\left(\frac{\lambda}{d}\right)
$$
As we just showed, the angle after the second grating will be $\theta_3=0$ again, but the ray will have translated by an amount given approximately by
$$
\delta x=L\ \theta_2=L\ \arcsin\left(\frac{\lambda}{d}\right),
$$
where $L$ is the distance between the two gratings.
This will achieve exactly what you desire since the point sources of different wavelengths will be imaged to different locations on your CCD.
Resolving Power
We can estimate the resolving power of such a spectrometer by considering two different closely spaced wavelengths: $\lambda_0$ and $\lambda_0+\delta\lambda$. In this case the separation on your CCD will be
$$
\begin{align}
\mu&=\delta x_2-\delta x_1\\
&=L\left[\arcsin\left(\frac{\lambda_0+\delta\lambda}{d}\right)-
\arcsin\left(\frac{\lambda_0}{d}\right)\right]\\
&\approx \frac{L\ \delta\lambda}{\sqrt{d^2-\lambda_0^2}}.
\end{align}
$$
So the resolving power will be
$$
\delta\lambda=\frac{\mu}{L}\sqrt{d^2-\lambda_0^2},
$$
where $\mu$ is the CCD pixel separation. Lets use a grating separation of $L=10\ \text{cm}$, a typical CCD pixel size is $\mu=20\ \mu\text{m}$, a CD has a diffraction grating spacing of $1.5\ \mu\text{m}$, and red light has a wavelength of $\lambda_0=600\ \text{nm}$. Putting all of these numbers in gives a resolving power at the CCD of
$$
\delta\lambda=0.25\ \text{nm}.
$$
In reality, you will probably hit some imperfections in a simple home setup which will prevent you from reaching the full sensitivity (grating imperfections, relative grating misalignment, etc), but it seems like $1\ \text{nm}$ is very achievable.
Best Answer
The resulting images from this spectrometer are promising.
A first starting point might to use something better than a cereal box. Preferably a box which is completely black inside, this improves the contrast of the image.
A second step could be to replace the CD by a diffraction grating and a mirror which are both not curved, the input slit could be replaced by something, where you can vary the distance with screws.
For any kind of 'real' science you might want to measure the output spectra. So instead of projecting it onto the nearest surface a fixed paper screen might be nice, where you can draw a scale for the individual wavelengths, e.g. to identify different elements (e.g., Na-lamps at night, fluorescent light bulbs).