Why is the voltage across a capacitor and the current across an inductor continuous?

If the voltage across a capacitor was discontinuous the charge on the capacitor would be discontinuous which would mean a transfer of charge in an infinitely short period of time i.e. an infinite current.

A similar argument applies for an inductor except this time a discontinuous change in the current would imply an infinite rate of change of current which for an inductor would mean an infinite induced emf opposing the change producing it.

Farcher's answer is correct, but I would like to expand a bit on a point that is frequently unclear to students.

Circuit theory deals with mathematical models of electric circuits and elements. For circuit theory, a capacitor is a two-terminal element defined by the constitutive relationship

$$i = C\frac{\mathrm{d} v}{\mathrm{d} t},$$

where $C$ is the capacitance, $v$ the voltage across the capacitor terminals, and $i$ the current crossing the capacitor.

In the definition above, there is something missing: what kind of functions are $v$ and $i$? The common, implicit, assumption is that $v$ and $i$ are real functions of real variable of whatever type needed for the above equation to make mathematical sense. In this case, if $v$ is differentiable at time $t$ – for the above equation to make sense – it should be continuous at $t$ too, for a a well-known theorem of mathematical analysis.

However, we can allow for possible discontinuous voltages if we extend the relationship to distributions. Such an extension can be convenient in certain cases, but of course one should be cautious in the interpretation of the results. Another generalization that is frequently employed in circuit theory is that of considering $v$ and $i$ as complex-valued functions (or distributions).