Bandwidth and throughput are fundamentally different concepts: the difference is much deeper than that between theory and practice. I shall take the meaning of 'throughput' as data-transfer capacity, measured in information per unit time, which I believe others will agree on.

Unfortunately, the original meaning of bandwidth has been deformed, possibly during the 1990s Internet marketing craze (see broadband). According to The Jargon File, the use of 'bandwidth' as a synonym for data throughput originated as an inside joke of hackers; it has since leaked into common usage as a supposedly proper synonym.

I acknowledge that languages evolve, and a word can come to mean something different over time. However, in the case of bandwidth the original meaning is alive and well, at least in science and engineering. For the sake of clarity we should remember the distinction.

Examples

Consider CD audio data. It is sampled with 16 bits at 44.1 kHz. This means a data rate of 705.6 kbps per one channel. On the other hand, Nyquist's theorem tells that the maximum bandwidth of CD audio is half of the sampling rate, i.e. 22.05 kHz.

The numbers 22.05 and 705.6 are different, but that does not explain much; bandwidth and throughput are measured in different units, so they cannot be meaningfully compared.

Turning this idea around, if we have an analogue channel with the bandwidth of 22.05 kHz, we could imagine using it to transfer data at 705.6 kbps. Just divide the data into 16-bit chunks and perform DAC.

That 16 bits converted into analogue means 216 = 65536 different signal levels. Each change in the signal level conveys 16 bits of information. This is the crux of the misunderstanding. Bandwidth basically tells how often you can change the signal level, while throughput also depends on the number of possible levels (Baud is not bps). Even a modest bandwidth can have a huge data throughput, given enough signal levels.

Which brings us to another example: Telephone lines have a bandwidth of only a few kHz. Yet it is possible to transfer data at 56 kbps. In this case, 256 signal levels are used. The number of signal levels is a compromise, because finer level separations mean more problems with noise. For example, a range of 0 to 5 volts divided into 256 levels means a separation of about 0.02 V, so the noise voltage must be less than this.

For a general mathematical relation between bandwidth and throughput, see Shannon's Law.

Conclusions

From the examples you may infer that throughput is proportional to bandwidth, other things being equal. It is because of these other things that bandwidth is not the same as throughput; for example, a different number of signal levels will change the constant of proportionality.

Another kind of change comes from the number of channels used. In our example of CD audio, one channel has a data rate of 705.6 kbps, so the typical setup of two channels (left and right) has 1411.2 kbps. Yet the bandwidth remains at 22.05 kHz, because the individual bandwidths do not add up in any well-defined way.

Interestingly enough, the term broadband does bear some relevance to home Internet services such as DSL and cable. It has to do with the details of the technology, AFAIK the kind of modulation used (please correct me if I'm wrong). However, the broadness does not directly refer to high data capacity, and it is physically quite possible to have a fat pipe (i.e. high data rate) without using broadband technology.