The choice of sequences matters.

The trick is to notice that if $X$ is such a space, then the copy of the rational numbers inside of $X$ -- let's denote it $\mathbb Q_X$ -- is a countable dense subset of $X$. This implies that if $X$ and $Y$ are any two such spaces, any homeomorphism $X \rightarrow Y$ is induced by a bijection $\mathbb Q_X \rightarrow \mathbb Q_Y$.

Using this observation, we can build, via a transfinite recursion of length $\mathfrak{c}$, two such spaces that are not homeomorphic. Begin with two copies $\mathbb R_X$ and $\mathbb R_Y$ of $\mathbb R$, and an enumeration $\langle f_\alpha :\, \alpha < \mathfrak{c} \rangle$ of all bijections $\mathbb Q_X \rightarrow \mathbb Q_Y$ (of which there are $\mathfrak{c}$-many, because $\mathbb Q_X$ and $\mathbb Q_Y$ are countable). At stage $\alpha$ of the recursion, suppose we've already chosen some sequences for your topologies on $X$ and $Y$, but fewer than $\mathfrak{c}$ sequences have been chosen in total. Then we can choose finitely many sequences of rationals converging to irrationals in $\mathbb R_X$ and $\mathbb R_Y$ in such a way that the map $f_\alpha$ cannot induce a homeomorphism between $X$ and $Y$. (In detail, either (1) $f_\alpha$ does not extend to a homeomorphism $\mathbb R_X \rightarrow \mathbb R_Y$, this is witnessed by a point whose sequences are not yet chosen, and we can choose a converging sequence in one space that does not map to a converging sequence in the other. Otherwise, (2) $f_\alpha$ does represent a homeomorphism of the reals, or at least it does on all the points we have left to consider, but we can choose a sequence converging in $X$ to some irrational $r$, and then choose in $Y$ a sequence converging to the image of $r$ under that homeomorphism that is disjoint from the image of the sequence we chose in $X$.) By the end of the recursion, we've chosen $\mathfrak{c}$ rational sequences for each space and killed all possible homeomorphisms between them. I didn't set up the recursion in such a way that we've necessarily chosen a sequence for every irrational yet, but the remaining choices (if there are any) can be made arbitrarily.