A function $f:X\to Y$ is continuous almost everywhere if it is only discontinuous on a set of measure $0$. Informally, this means that if we choose a random point on the function, the probability that it is continuous is exactly $1$.
For example, any countable set, such as $\mathbb Q$, will be of measure $0$ in an uncountable one, such as $\mathbb R$.
There is an important distinction to make here: a function is continuous almost everywhere if it is continuous on a "large" subset. But it must be continuous on that subset. Your function $g$ is continuous at every irrational point, and hence we can say that it is continuous almost everywhere.
That is not the same as removing a countable set of values from the function, because in doing so, you can damage the continuity of the remaining points. In the case of $f$, you are left with a function that is nowhere continuous!
This only works if you remove a countable set, but make sure that the remaining points are still continuous.
First bullet point is correct.
Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.
Hint for third: Very similar construction to the 4th, which is correct.
Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.
To do that, consider such a point $x_0$ and a given $\epsilon > 0$. We know that because of the uniform convergence of $f_n$ to $f$, there is a $N \in \mathbb N$ such that
$$ \forall x \in [a,b], \forall n > N: |f_n(x) - f(x)| < \frac {\epsilon}3$$
We also know that $f_{N+1}$ is continuous at $x_0$, which means
$$ \exists \delta > 0: |x-x_0|<\delta \Rightarrow |f_{N+1}(x) - f_{N+1} (x_0)| < \frac {\epsilon}3$$
Together this means
$$ |x-x_0|<\delta \Rightarrow |f(x) - f(x_0)| = |f(x) - f_{N+1}(x) + f_{N+1}(x) - f_{N+1}(x_0)+ f_{N+1}(x_0) -f(x_0)| \le |f(x) - f_{N+1}(x)| + |f_{N+1}(x) - f_{N+1}(x_0)| + |f_{N+1}(x_0) -f(x_0)|< \frac {\epsilon}3 + \frac {\epsilon}3 + \frac {\epsilon}3 < \epsilon$$
In a nutshell, this is the proof that uniform convergence of a function sequence of continuous functions implies continuity of the limit.
For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.
Best Answer
There is not. Suppose $g\colon X\rightarrow\mathbb{R}$ is such a function. $f^{-1}(0)$ has positive measure and $f=g$ a.e., so $g^{-1}(0)$ has positive measure; similarly, $g^{-1}(1)$ has positive measure. In particular, $0,1\in g(X)$, so $[0,1]\subseteq g(X)$ by the IVT. Now, the intuition is that the "jump" from $0$ to $1$ that $f$ does is something that a continuous $g$ could only do on a set with positive measure. Indeed, $(0,1)$ is an open set, so $g^{-1}((0,1))$ is open by continuity of $g$, non-empty by the IVT and hence has positive measure. However $f(x)\neq g(x)$ for all $x\in g^{-1}((0,1))$, because $f(X)=\{0,1\}$. Hence such a $g$ does not exist. I leave it to you to generalize this to arbitrary jump discontinuities.