The one-dimensional time-independent Schroedinger Equation is defined as
$$E\phi(x)=-\frac{\hbar^2}{2m}\frac{d^2\phi(x)}{dx^2}+V(x)\phi(x)$$
In my QM1 lecture, the Professor said that the solutions $\phi(x)$ are continuous because the Schroedinger Equation is a differential Equation. Why is this the case? Which theorem states that the solutions to differential equations are continuous?
On
Chicken or egg?
It is the other way round: one writes differential equations for the quantities described by continuous functions. At the discontinuities the derivative of the functions (at least some of them) are poorly defined, which may render the equation involving these derivatives meaningless. Thus, Schrödinger equation implies that the wave functions are at least twice differentiable.
Everyday life (of a physicist)
In practice one does work with discontinuous functions, which is either a bit of cheating or an approximation grounded in the more sophisticated theory of generalized functions. The most-encountered example is the delta-function potential, where the derivative of the wave function experiences a jump. This could be justified using the theory of generalized functions or by considering a very narrow rectangular well/barrier. (Note that the discontinuity of the potential does not cause trouble, as it is not differentiated - it simply translates into the boundary conditions.)
If you want a solution to the equation as stated, the second derivative must exist for the equation to have any meaning. As a result the first derivative must exist and so the function must be continuous.
However, we are sometimes interested in potentials that act like a delta distribution or potentials that are infinite in such a way as to preclude finding a solution for which a finite second derivative exists at all points. In these cases, we could either give up or weaken our initial requirement. One way to do so would be to integrate both sides of the equation and only require that the integral of the left hand side equals the integral of the right. Then we only need the first derivative to exist and so we can find pick potentials which would result in solutions with discontinuous first derivatives. (Try plugging in a delta distribution into this integrated eigenvalue problem)
Another way we could weaken the requirement is by defining a sequence of potentials, which in the limit approaches our badly behaved potential of interest. For each potential in the sequence we can find a solution in the strict sense and then we can ask what the solution looks like in the limit. This limit solution need not have a second derivative. In fact, you can find a sequence of potentials, which generates a step function as a solution in the limit. (Try it! pick a limit of functions which approaches a step function, differentiate twice and figure out what potential would result in the sequence of solutions)
One more thought: it is possible to make a physical argument that rules out an infinite first derivative and therefore also a discontinuous solution. If the first derivative is infinite, then the momentum is, too. That implies the kinetic energy is infinite. Although, I'm not sure if it also implies the total energy is infinite. In principle, I'm not sure if infinite kinetic energy is a problem as long as the total energy is finite in the limit.