The proof that $\text{Ext} \neq \omega$ is given below. However, to address you qustion, $\text{Ext}$ is just the set of $i$ such that $\varphi_i(x)$, which may not be total can be extending to a total computable function $\varphi_j$ such that for all $x$ such that $\varphi_i(x)\downarrow$, $\varphi_j(x) \downarrow = \varphi_i(x)$.
Also it is not true that $i \in \text{Ext}$ if there is a $j$ such that $W_i \subset W_j$ and $W_j$ is computable. For example, every $W_i$ is a subset of $\omega$, which is computable. This would imply $\text{Ext} = \omega$, which you are asked to prove is not the case.
Also, you said that $W_i \subset W_j$ where $W_j$ is computable implies $W_i$ is computable. Again, every c.e. set is a subset of $\omega$ which is computable. If your statement was true then every set would be computable. Clearly, $K$ the halting problem is not.
By the enumeration theorem, there exists a index $z$ such that
$\varphi_z(e,x) = \varphi_e(x)$
That is, $z$ is the index for the universal function.
Let $k$ be the index of the following partial computable function.
$\varphi_k(x) = \varphi_z(x,x)$
The claim is that $k \notin \text{Ext}$. Suppose $k \in \text{Ext}$, this would imply that $\varphi_k$ can be extended to a total computable function $\varphi'(x)$. Now define the computable function $\theta$ as follow :
$\theta(x) = \varphi'(x) + 1$
Clearly $\theta$ is a partial (actually total) computable function. Therefore, it has an index. However, it can not be $\varphi_x$ for any any $x$. This is because if $\varphi_x(x)$ converges, then $\theta(x) = \varphi'(x) + 1 = \varphi_z(x,x) + 1 = \varphi_x(x) + 1 \neq \varphi_x(x)$. If $\varphi_x(x)$ does not coverges, then it can not be equal to $\theta(x)$ since $\theta$ is a total computable function. $\theta \neq \varphi_x$. Contradiction since $\theta$ has no index. Thus $k \notin \text{Ext}$. $\text{Ext} \neq \omega$.
At least in Soare's new book, he asks this question twice. The second you are mean to use the computably inseparable thing. If I can remember how to do it that way I will post a proof later.
Here is the proof of the result using the computably inseparable pair. Recall that $A$ and $B$ are computably inseparable if there does not exists a computable set $C$ such that $A \subset C$ and $A \cap C = \emptyset$. I assumed that you proved already that there exists an inseparable pair $A$ and $B$, where $A$ and $B$ are c.e.
Define
$\psi(x) = \begin{cases}
1 & \quad x \in A \\
0 & \quad x \in B
\end{cases}$
This function partial computable (it diverges if $x$ is in neither). Therefore, this function $\psi$ has an index, say $e$. Assuming $\text{Ext} = \omega$, $\varphi_e$ can be extended to a total computable function $\varphi_f$. Since $\varphi_f$ is total and $0,1$ valued, let $C$ be $\{x : \varphi_f(x) = 1\}$, which is computable since $\varphi_f$ is total. Clearly, $A \subset C$ since $\varphi_f(x) = 1$ whenever $\varphi_e(x) = 1$ and $\varphi_e(x) = 1$ if and only if $x \in A$. Similarly, $A \cap C = \emptyset$ since $\varphi_f(x) = 0$ if $\varphi_e(x) = 0$ and $\varphi_e(x) = 0$ if and only if $x \in B$. Thus $C$ contradict that fact that $A$ and $B$ are computable inseparable. Contradiction!
This function is computable. Clearly (by the definition of $g$) we have $g(x) = 1$ for all $x \geq 1$. Also $g(0)$ can be either $0$ or $1$. Regardless, this function is computable: in the latter case it is the constant function $g(x) \equiv 1$ and in the first case it is the charateristic function of the set of positive numbers $g(x) = \text{sg}(x)$, which is recursive.
Alternatively, this function differs from (trivially computable) the constant $1$ function at most in one point $x = 0$. It is quite easy to show that if you change the values of some computable function in finite number of points you obtain the function which is also computable.
This exercise shows that trivially computable functions can be defined using some intricate set of instructions. Another well-known example of such a function is:
$$f(x) = \begin{cases}
1, & \text{if a consecutive run of at least $x$ $9$'s occurs in the decimal expansion of $\pi$,}\\
0, & \text{otherwise.}
\end{cases}$$
It is either the constant $1$ function or equals $1$ at the finite initial segment of $\mathbb{N}$ and hence differs from the constant $0$ function only at the finite number of points.
One more example:
$$f(x) = \begin{cases}
0, & \text{if $\mathsf{ZFC} \vdash P \neq NP$,}\\
1, & \text{if $\mathsf{ZFC} \vdash P = NP,$}\\
2, & \text{otherwise.}
\end{cases}$$
We don't know which of the cases takes place, but in any of them the resulting function is computable. At the present time we don't have an algorithm to compute this function, but it is still computable. So all these proofs are non-constructive.
Best Answer
In general, consider sets $A, B$ and functions $f : S \to B$, $g : R \to B$ where $S, R \subseteq A$. Equivalently, $f$ and $g$ are partial functions $A \to B$. Then we say “$g$ extends $f$” if and only if $S \subseteq R$ and $\forall s \in S (f(s) = g(s))$.
In this context, we have a partial recursive function - that is, a recursive function $f : S \to \mathbb{N}$ where $S \subseteq \mathbb{N}$. We are asking whether there is any total recursive function $g : \mathbb{N} \to \mathbb{N}$ such that $\forall s \in S (g(s) = f(s))$.
The example your classmate provided is valid. For let us suppose $\psi$ extends to a total recursive function $\theta : \mathbb{N} \to \mathbb{N}$. Then I claim that the set $\{x \in \mathbb{N} \mid \phi_x(x)$ is defined $\}$ is recursively decidable.
For if we wish to decide whether $\phi_x(x)$ is defined, we first compute $\theta(x)$. We then see whether the computation of $\phi_x(x)$ halts in exactly $\theta(x)$ steps. If so, then clearly $\phi_x(x)$ is defined. If not, then suppose $\phi_x(x)$ were defined; then the computation of $\phi_x(x)$ would halt in $n$ steps for some $n$. That is, $\psi(x) = n$. Since $\psi(x)$ is defined, $\psi(x) = \theta(x)$. So we would have that $\phi_x(x)$ takes exactly $\theta(x)$ steps to compute.
But it is well-known that $\{x \mid \phi_x(x)$ is defined$\}$ cannot be recursively decidable. Therefore, $\psi$ can’t be extended.