Given a bounded countable set, for all uncountable sets, does part of the uncountable set look the exact same as that countable set?

Given an uncountable set $\ X\subset \mathbb{R}\ $ and a bounded countable set $\ Y\subset\mathbb{R},\ $ does there exist $\ a,b\in\mathbb{R}\ $ such that $\ Y\subset \{a+bx:x\in X\}\ ?$

I was thinking that this could be true, because I know that an uncountable set must have condensation points, and it feels like, close to these condensation points, there should be points of many different spacings.

Update: Lubin in the comments gave the answer: $\ X =\ $ The standard Cantor set, $\ Y = [0,1] \cap \mathbb{Q},\ $ which I think answers the question.

But what if we don't allow $\ Y\ $ to be dense anywhere? The answer is still probably no, but this might be due to some interesting classification of countable and uncountable sets.

A counterexample exists for arbitrary $Y$ with even just $3$ points. That is, given any $Y\subseteq\mathbb{R}$ at least $3$ points, there exists an uncountable $X\subseteq\mathbb{R}$ (in fact, an $X$ of cardinality $2^{\aleph_0}$) which does not contain any translated and scaled copy of $Y$.

The proof is just to pick the points of $X$ one by one by transfinite recursion. We may assume that $Y$ has exactly $3$ points, say $Y=\{a,b,c\}$ (if $Y$ has more points, just take a $3$-point subset of it). Suppose we have a set $S\subseteq \mathbb{R}$ of cardinality less than $2^{\aleph_0}$ which contains no translated and scaled copy of $Y$. Then I claim we can always add one more point to $S$ to preserve this property. Indeed, for any two points $s,t\in S$, if we were to take a copy of $Y$ with $s$ corresponding to $a$ and $t$ corresponding to $b$, there is only one point that could then correspond to $c$, and similarly for all the other ways $s$ and $t$ could correspond to two points of $Y$. So for any pair of points in $S$, there are only finitely many points of $\mathbb{R}$ which we are forbidden from adding to $S$ since they would form a copy of $Y$ with those two points. In total, then, since there are fewer than $2^{\aleph_0}$ pairs of points of $S$, there are fewer than $2^{\aleph_0}$ points of $\mathbb{R}$ which would form a copy of $Y$ together with two points of $S$. So, if we take a point of $\mathbb{R}$ that is not one of those forbidden points and add it to $S$, the resulting set still contains no copy of $Y$.

So, we can construct a sequence of distinct points $(x_\alpha)_{\alpha<\mathfrak{c}}$ which contains no copy of $Y$ by transfinite recursion. At each step, we have chosen fewer than $2^{\aleph_0}$ points so far, so we can pick one more point such that our points still do not contain any copy of $Y$. At limit ordinals, we cannot gain a copy of $Y$, since $Y$ is finite so any copy would have been contained at an earlier step already. In the end, the set $X=\{x_\alpha:\alpha<\mathfrak{c}\}$ has cardinality $2^{\aleph_0}$ and does not contain any copy of $Y$.