It looks like no.

Assume the contrary. We may start with two distinct rationals $q_1,p_1$ such that the sets $f(q_1\times \mathbb{R}^{w-1})=f(\{(q_1,\cdot,\cdot,\dots)\})$ and $f(p_1\times \mathbb{R}^{w-1})$ intersect. Indeed, for any $p_1$ the set $f(p_1\times \mathbb{R}^{w-1})$ has non-empty interior, choose reals such that $f(p_1,x_2,x_3,\dots)$ is interior point of this set. After that any $q_1$ close enough to $p_1$ works, since $f(q_1,x_2,\dots)$ is close to $f(p_1,x_2,\dots)$.

After that choose small intervals $\Delta_2,\Delta_3,\dots$ and $\delta_2,\delta_3,\dots$ such that the sets $f(q_1\times \Delta_2\times \Delta_3\times \dots)$ and $f(p_1\times \delta_2\times \delta_3\times \dots)$ intersect and have diameter less than 1. After that choose $q_2\in \Delta_2$ and $p_2\in \delta_2$ such that $f(q_1\times q_2\times \Delta_3\times \dots)$ and $f(p_1\times p_2\times \delta_3\times \dots)$ intersect. Proceeding this way, we get two rational sequences with the same value of $f$.

It looks like no.

Assume the contrary. We may start with two distinct rationals $q_1,p_1$ such that the sets $f(q_1\times \mathbb{R}^{\omega-1})=f(\{(q_1,\cdot,\cdot,\dots)\})$ and $f(p_1\times \mathbb{R}^{\omega-1})$ intersect. Indeed, for any $p_1$ the set $f(p_1\times \mathbb{R}^{\omega-1})$ has non-empty interior, choose reals such that $f(p_1,x_2,x_3,\dots)$ is interior point of this set. After that any $q_1$ close enough to $p_1$ works, since $f(q_1,x_2,\dots)$ is close to $f(p_1,x_2,\dots)$.

After that choose small intervals $\Delta_2,\Delta_3,\dots$ and $\delta_2,\delta_3,\dots$ such that the sets $f(q_1\times \Delta_2\times \Delta_3\times \dots)$ and $f(p_1\times \delta_2\times \delta_3\times \dots)$ intersect and have diameter less than 1. After that choose $q_2\in \Delta_2$ and $p_2\in \delta_2$ such that $f(q_1\times q_2\times \Delta_3\times \dots)$ and $f(p_1\times p_2\times \delta_3\times \dots)$ intersect. Proceeding this way, we get two rational sequences with the same value of $f$.

It looks like no.

Assume the contrary. We may start with two distinct rationals $q_1,p_1$ such that the sets $f(q_1\times \mathbb{R}^{w-1})=f(\{(q_1,\cdot,\cdot,\dots)\})$ and $f(p_1\times \mathbb{R}^{w-1})$ intersect. Indeed, for any $q_1$ the set $f(p_1\times \mathbb{R}^{w-1})$ has non-empty interior, choose reals such that $f(p_1,x_2,x_3,\dots)$ is interior point of this set. After that any $q_1$ close enough to $p_1$ works, since $f(q_1,x_2,\dots)$ is close to $f(p_1,x_2,\dots)$.

After that choose small intervals $\Delta_2,\Delta_3,\dots$ and $\delta_2,\delta_3,\dots$ such that the sets $f(q_1\times \Delta_2\times \Delta_3\times \dots)$ and $f(p_1\times \delta_2\times \delta_3\times \dots)$ intersect and have diameter less than 1. After that choose $q_2\in \Delta_2$ and $p_2\in \delta_2$ such that $f(q_1\times q_2\times \Delta_3\times \dots)$ and $f(p_1\times p_2\times \delta_3\times \dots)$ intersect. Proceeding this way, we get two rational sequences with the same value of $f$.

It looks like no.

Assume the contrary. We may start with two distinct rationals $q_1,p_1$ such that the sets $f(q_1\times \mathbb{R}^{w-1})=f(\{(q_1,\cdot,\cdot,\dots)\})$ and $f(p_1\times \mathbb{R}^{w-1})$ intersect. Indeed, for any $p_1$ the set $f(p_1\times \mathbb{R}^{w-1})$ has non-empty interior, choose reals such that $f(p_1,x_2,x_3,\dots)$ is interior point of this set. After that any $q_1$ close enough to $p_1$ works, since $f(q_1,x_2,\dots)$ is close to $f(p_1,x_2,\dots)$.

After that choose small intervals $\Delta_2,\Delta_3,\dots$ and $\delta_2,\delta_3,\dots$ such that the sets $f(q_1\times \Delta_2\times \Delta_3\times \dots)$ and $f(p_1\times \delta_2\times \delta_3\times \dots)$ intersect and have diameter less than 1. After that choose $q_2\in \Delta_2$ and $p_2\in \delta_2$ such that $f(q_1\times q_2\times \Delta_3\times \dots)$ and $f(p_1\times p_2\times \delta_3\times \dots)$ intersect. Proceeding this way, we get two rational sequences with the same value of $f$.