Halting problem without input?
Here is a different proof using Berry's paradox instead of the liar paradox. Suppose that the halting problem (for programs with no input) were decidable. For each $n$, let $f(n)$ be the smallest number which cannot be computed by a program of length at most $n$ (without inputs). Consider the following program for computing $f(n)$:


  Go over all programs of length at most $n$, and for each one that halts, store the number it computes. Output the smallest number not encountered in this way.


This is computable since we can solve the halting problem for the relevant programs, and so we can determine which programs halt. What is the length of this program? The code itself is fixed apart from the length $n$, which can be encoded using $2\log n$ bits. All in all, this program has length $2\log n + C$ for some constant $C$. For large enough $n$ we have $2\log n + C \leq n$, and so there is a program of length at most $n$ which computes $f(n)$! This cannot be, and we are forced to conclude that the halting program is not computable.