Finite sequence input to DTFT
I doubt that Hilmar will agree (maybe he will) but I will offer this answer as a counter to the one that has already been (prematurely) accepted by the questioner.

Hilmar's answer is wrong.  It is factually wrong and analytically wrong.


  
    1) Does the DTFT take only infinite input sequences?
  
  
  No. There is nothing that prevents you from applying the formula to a finite sequence.


That's sorta true but it must be said that the "formula" is explicitly of an infinite sequence.  However, there is nothing in that formula that says any particular sample must take on any particular value (like non-zero) other than a finite value.  There is nothing that prevents all but a finite subset of samples having zero value.

The DTFT is:

$$ X(e^{j\omega}) = \sum\limits_{n=-\infty}^{\infty} x[n] \, e^{-j\omega n} $$

That's what the formula is and all of the $x[n]$ have a defined value, even if it's zero.


  However, a finite time sequence has an infinite spectrum, so you can't sample it without some amount of aliasing


No one is sampling it (again).  It has already been sampled.  The question is: What happens if the length of the sequence (which is, in the DTFT, infinite in length) is shortened to a finite length?

And the answer is: It depends on the manner in which the sequence is shortened to a finite length.  Or, if the sequence was never infinite in length and later applied to the DTFT, how are the infinite number of samples outside of the original finite length defined.

But if you do define all of those samples outside of your original finite length to be zero, that is saying:

$$ x[n] = 0 \qquad \text{for } x&lt;0 \text{ and } x\ge N &lt; \infty  $$

Then the DFT is equivalent to:

$$ X(e^{j\omega}) = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j\omega n} $$

And then it starts to look like the DFT.  If $X(e^{j\omega})$ is sampled at $N$ equally-spaced values on the unit circle including $X(e^{j0})$, then it is equivalent to the DFT

$$ X[k] = X(e^{j\omega}) \Big|_{\omega = \frac{2 \pi k}{N}} = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j2 \pi k n/N} $$

But something misleading in Hilmar's answer must be corrected:

Applying a finite-length sequence to the DTFT has nothing to do with aliasing.  Nothing at all.