We will show that Rice's theorem and the the halting problem are equivalent.

The Halting theorem implies Rice's theorem

Let $S$ be a non trivial set of computable partial functions, and suppose that there exists a Turing machine encoded by $\left[n\right]$ such that: $$\left[n\right]\left(m\right)=\{\begin{array}{cc}1& \left[m\right]\text{computes a function in}S\\ 0& \text{otherwise}\end{array}$$

We can assume w.l.o.g. that the empty function undefined on every input is not in $S$ ^[1]. Thus there exists a computable function in $S$ computed by some machine $\left[s\right]$ such that $\left[s\right]\left(x\right)$ is defined for some input $x$.

Suppose we want to decide whether the machine $\left[m\right]$ halts on input $\left[x\right]$.

For that purpose we can build a machine $Prox{y}_s$ which does the following:

Proxy_s(z):
    call [m](x)
    return s[z]

Clearly, if $\left[m\right]\left(x\right)$ halts then Proxy_z computes the same function as $\left[s\right]$, and thus $\left[n\right]\left(Prox{y}_s\right)=1$.

If on the other hand $\left[m\right]\left(x\right)$ does not halt, then Proxy_s(z) computes the empty function, which we assumed to not be in $S$, and therefore $\left[n\right]\left(Prox{y}_s\right)=0$.

Thus we can use a Turing machine computing pertinence to $S$ to decide the halting problem, which we know is undecidable. We conclude that such a machine cannot possibly exists, and thus Rice's theorem holds.

Rice's Theorem implies the Halting theorem

Suppose that there was a Turing machine $HALT$ deciding the Halting Problem.

Let $S$ be the set of computable functions defined on a fixed input $x$, which is clearly non-trivial, as it does not contain the empty function but is not empty either. Let $\left[n\right]$ be a Turing machine, and we want to decide whether $\left[n\right]\in S$ or not. If this was possible for an arbitrary $\left[n\right]$, then we would have reached a contradiction, as Rice's theorem forbids this outcome.

But $\left[n\right]$ belongs to $S$ iff $\left[n\right]$ halts on input $x$, so we can use $HALT$ to decide whether $\left[n\right]$ belongs to $S$, in contradiction with Rice's theorem. So our supposition of the existence of $HALT$ was erroneous, and thus the Halting theorem is true.

1. ^{^︎}

   Suppose that the empty function is in $S$. Then it is satisfied that the empty function is not in $S^c$, and if $S$ is decidable then it follows immediately that $S^c$ is decidable as well. So we can use $S^c$ as our $S$ and the argument follows exactly the same.