Construction of a model of ZF with a discontinuous but sequentially continuous function.

Well, if your sequence of rational numbers is in $M$, then $f$ is in $M$, since it was pretty straightforward to define from the sequence. If $f$ is not in $M$, then your forcing is not going to be in $M$ either, in which case, in what sense do we use it to force over $M$? Moreover, adding this sequence to $M$ will witness that $\alpha$ is countable, so you must have added ordinals, if you were to have a model of $\sf ZF$, which you cannot do with forcing, and quite possible, you might not be able to do it at all.

The idea, instead, is to consider Cohen's first model. This model is obtained by starting with a model of $\sf ZFC$, adding an $\omega$-sequence of Cohen reals, and then using permutations to "forget" the sequence, and only remember the set of these Cohen reals. The result is that this set is Dedekind-finite, so it does not contain any nontrivial sequences.

Call this set $A$, and now pick some $a\in A$ and consider the function, defined on $A$, which is $1$ everywhere, except $a$, and $0$ otherwise. Since all convergent sequences in $A$ are eventually constant, this is trivially sequentially continuous; and it is not hard to show that $A$ must be dense, and therefore the function is not continuous at $a$.