I think one important thing that you must understand is that a Dirac delta impulse is not an ordinary function that can be evaluated. So you do not multiply your signal with a value of $\infty$. What happens instead is that you weigh the individual shifted Dirac impulses with the corresponding values of the signal due to the following identity:
which is true as long as $x(t)$ is continuous at $t=nT$.
Consequently, multiplying a signal with a Dirac impulse train results in a weighted impulse train, where the weights are the signal values at the sample instants. So what happens is that from a continuous signal $x(t)$ you only retain the sample values $x(nT)$, but you still have an expression that can be considered a continuous-time signal (in the sense that it can be integrated or convolved with another function).
Note that this is just a mathematical model. As pointed out in Carlos Danger's answer, what usually happens is that this impulse train is filtered, i.e., you get
where $h(t)$ is the impulse response of the filter. Since $\delta(t-nT)\star h(t)=h(t-nT)$, Eq. $(2)$ equals
which is now an ordinary function which can be evaluated for any $t$.