Where is the error in this logic for halting Turing machines?

Question

Let $\mathbb{H}$ be the set of all Turing machines that halt on all inputs. Consider the following Turing machine $T$. On input $\langle S \rangle$ where $S \in \mathbb{H}$ (note that the angle brackets indicate a string encoding of $S$), $T$ does the following:

1. Simulate $S$ on $\langle S \rangle$.
2. If $S$ accepts, reject. If $S$ rejects, accept.

Because $S \in \mathbb{H}$, $S$ will always halt. Thus, $T$ will always halt, which implies $T \in \mathbb{H}$. Note that $T$ decides the language

$$\{\langle S \rangle \mid \text{$S \in \mathbb{H}$ and $S$ rejects $\langle S \rangle$}\}.$$

But $T$ accepts $\langle T \rangle$ if and only if $T$ rejects $\langle T \rangle$—this is a contradiction. Where is the error in this logic?

Answer 1

It is false that $T$ decides the language $$L = \{\langle S \rangle \mid \text{$S \in \mathbb{H}$ and $S$ rejects $\langle S \rangle$}\}.$$

In particular consider a turing machine $M$ that does not halt on all inputs but halts and rejects on input $\langle M \rangle$. Then $T(\langle M \rangle)$ accepts but $\langle M \rangle$ is not in $L$ since $M \not\in \mathbb{H}$.

Answer 2

I just realized where I went wrong. The Turing machine $T$ that I gave does not actually exist.

In particular, I required that $S \in \mathbb{H}$ for an input $\langle S \rangle$. But this implicitly tasks $T$ with verifying that $S \in \mathbb{H}$ before executing steps 1 and 2. This requires $T$ to solve the halting problem (or some variant of it).