Let $X_n,Y_n$ be two sequences of random variable, where $0<X_n<Y_n$. Does $Y_n=O_p(1)$ imply $X_n=O_p(1)?$

Let $X_n,Y_n$ be two sequences of random variable, where $0<X_n<Y_n$. Does $Y_n=O_p(1)$ imply $X_n=O_p(1)?$

My proof

Since $Y_n=O_p(1)$, $$ (\forall \epsilon>0)(\exists k)\Big(P(|Y_n|>k)\leq \epsilon\Big)\ \ $$ Next, by law of total probability $$ P(|X_n|>k)=\underbrace{P(|Y_n|>k)}_{\leq \epsilon}\underbrace{P(|X_n|>k|\ |Y_n|>k)}_{<1}+\underbrace{P(|Y_n|\leq k)}_{\geq 1-\epsilon }\underbrace{P(|X_n|>k|\ |Y_n|\leq k)}_{0}\leq \epsilon $$ So the same $k$ works.

Is my proof correct? I will also accept any alternative proof (or counter proof) as answer.

What you did is correct. I think that you already saw the fact that if $A\subset B$, then $\mathbb P(A)\leqslant \mathbb P(B)$ (which is also a consequence of the law of total probability) hence applying this to $A=\left\{\lvert X_n\rvert>k\right\}$ and $B=\left\{\lvert Y_n\rvert>k\right\}$ would directly give that $\mathbb P\left(\left\{\lvert Y_n\rvert>k\right\}\right)<\varepsilon$.