Let $X_1,...,X_n \sim \mathrm{Poisson}(\lambda_n)$ where $\lambda_n=1/n$ and let $n \in \mathbb{N}$. I have to show that $Y_n=nX_n$ converges to $0$ in probability. My approach: (using Markov's inequality)
Note that $Y_n$ is a non-negative random variable since $n\geq 1$ and $X_n$ is also non-negative (Poisson). Therefore by Markov's inequality it holds that:
$$P(Y_n >t) \leq \frac{E(Y_n)}{t} = \frac{E(nX_n)}{t}=\frac{nE(X_n)}{t} = \frac{n\lambda_n}{t} = \frac{1}{t},$$
for all $t\in \mathbb{R}_{>0}$. However for $n \rightarrow \infty$ we do not see that $P(Y_n > t)\rightarrow 0$ so what is wrong with this approach?
Just compute directly. For any $\epsilon>0$, $$ P(|Y_n|>\epsilon) =P(nX_n>\epsilon) =P(X_n>\epsilon/n) \le P(X_n> 0)=1-e^{-1/n}\stackrel{n\to\infty}\longrightarrow 0. $$ There is nothing much to say about what is wrong with your approach. It was a good idea, you executed it well, but it turned out not to be useful. That is just how it goes sometimes in math!