Are $\mathbb{Q}$ and $X=\left\{1,\frac{1}{2}, \frac{1}{3},…\right\}$ homeomorph?

Determine whether $\mathbb{Q}$ and $X=\left\{1,\frac{1}{2}, \frac{1}{3},...\right\}$ are homeomorph, where each set is equipped with the subspace topology induced by $\mathbb{R}$.

I'm having trouble with the last part. Are these the two topologies? $$T_{\mathbb{Q}}=\left\{ \mathbb{Q} \cap U; U \text{ open in } \mathbb{R} \right\} \\T_{X}=\left\{ X \cap U; U \text{ open in } \mathbb{R} \right\}$$

It would appear that $\mathbb{Q}$ and $X=\left\{1,\frac{1}{2}, \frac{1}{3},...\right\}$ are homeomorph, but I'm not sure. I've tried to find a homeomorphism, but I haven't managed to do so.

Any hint would be greatly appreciated.

Yes, $T_{\mathbb Q}$ and $T_X$ are the respective subspace topologies.
And no, $\mathbb Q$ and $X$ are not homeomorphic, because in $X$ every singleton set is open (i.e. $X$ is discrete) whereas in $\mathbb Q$, no singleton set is open.
Hint: $X$ has isolated points.