About computable sets
Let $T$ be the set of all Turing machines.

Let $B = \{ (e,x, y) \in T \times \{0,1\}^* \times \mathbb{N} : e(x) \text{ halts in $y$ steps}\}$, and define $C = \{e \in T : \forall x \in \{0,1\}^*, \; \exists y \in \mathbb{N}, \; (e, x, y) \in B]$.

We start by proving $\text{TOT}=C$.

If $e \in \text{TOT}$ then, $\forall x \in \{0,1\}^*$, $e(x)$ halts. Let $y_x$ be the number of steps needed for $e(x)$ to halt. Then $ \forall x \in \{0,1\}^*$ there is a choice of $y \in \mathbb{N}$, namely $y = y_x$, such that $(e,x,y) \in B$, showing that $e \in C$ and hence $\text{TOT} \subseteq C$.

If $e \not\in \text{TOT}$, then $\exists x \in \{0,1\}^*$ such that $e(x)$ does not halt. Then $\exists x$, such that $\forall y \in \mathbb{N}$, $(e, x, y) \not\in B$, showing that if is false that $\forall x \in \{0,1\}^*, \exists y \in \mathbb{N}$ such that $(e, x, y) \in B$. As a consequence $e \not\in C$ implying $\text{TOT} \supseteq C$.

It remains to show that $B$ is computable.
To decide whether any given triple $(e, x,y) \in T \times \{0,1\}^* \times \mathbb{N}$ belongs to $B$  it suffices to consider a Turing machine $T'$ that simulates $e(x)$ for $y$ steps. If $e(x)$ halts, then $T'$ accepts. Otherwise, if $e(x)$ does not halt by the end of the $y$-th step, $T'$ rejects.