Proof (confusion) of the product of two random variables of two sequences converge to XY

It is not a matter of searching such $K$ but of knowing that such $K$ exists. That is expressed in words like: "we can find/choose a $K$ such that..."

In stead of answering your question I will give a setup that might be less confusing.

For every $\epsilon>0$ and $k>0$ and positive integer $n$ it is true for $x_{n},y_{n}x,y\in\mathbb{R}$ that:

$\begin{aligned}\left|x_{n}-x\right|\leq\frac{\epsilon}{4k}\wedge\left|y_{n}\right|\leq2k\wedge\left|y_{n}-y\right|\leq\frac{\epsilon}{2k}\wedge\left|x\right|\leq k & \implies\left|x_{n}-x\right|\left|y_{n}\right|+\left|y_{n}-y\right|\left|x\right|\leq\epsilon\\ & \implies\left|\left(x_{n}-x\right)y_{n}\right|+\left|\left(y_{n}-y\right)x\right|\leq\epsilon\\ & \implies\left|\left(x_{n}-x\right)y_{n}+\left(y_{n}-y\right)x\right|\leq\epsilon\\ & \implies\left|x_{n}y_{n}-xy\right|\leq\epsilon \end{aligned} $

and also that:

$\begin{aligned}\left|y_{n}-y\right|\leq k\wedge\left|y\right|\leq k & \implies\left|y_{n}-y\right|+\left|y\right|\leq2k\\ & \implies\left|\left(y_{n}-y\right)+y\right|\leq2k\\ & \implies\left|y_{n}\right|\leq2k \end{aligned} $

Turning the inequalities around we find equivalently:

$\left|x_{n}y_{n}-xy\right|>\epsilon\implies\left|x_{n}-x\right|>\frac{\epsilon}{4k}\vee\left|y_{n}\right|>2k\vee\left|y_{n}-y\right|>\frac{\epsilon}{2k}\vee\left|x\right|>k$

and

$\left|y_{n}\right|>2k\implies\left|y_{n}-y\right|>k\vee\left|y\right|>k$

justifying the inequalities:

$$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq$$$$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{4k}\right)+\mathsf{P}\left(\left|Y_{n}\right|>2k\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{2k}\right)+\mathsf{P}\left(\left|X\right|>k\right)$$

and

$$\mathsf{P}\left(\left|Y_{n}\right|>2k\right)\leq\mathsf{P}\left(\left|Y_{n}-Y\right|>k\right)+\mathsf{P}\left(\left|Y\right|>k\right)$$

Combining both inequalities we find:

$$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq$$$$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{4k}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>k\right)+\mathsf{P}\left(\left|Y\right|>k\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{2k}\right)+\mathsf{P}\left(\left|X\right|>k\right)$$

For a fixed $\delta>0$ a $k_{\delta}>0$ exists with $\mathsf{P}\left(\left|Y\right|>k_{\delta}\right)<\frac{1}{3}\delta$ and $\mathsf{P}\left(\left|X\right|>k_{\delta}\right)<\frac{1}{3}\delta$

Substituting $k=k_{\delta}$ (which is allowed!) we find:

$$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq$$$$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{4k_{\delta}}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>k_{\delta}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{2k_{\delta}}\right)+\frac{2}{3}\delta$$

Since $X_{n}\stackrel{p}{\to}X$ and $Y_{n}\stackrel{p}{\to}Y$ some $n_{\delta}$ exists with $$\mathsf{P}\left(\left|X_{n}-X\right|>\frac{\epsilon}{2k_{\delta}}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>k_{\delta}\right)+\mathsf{P}\left(\left|Y_{n}-Y\right|>\frac{\epsilon}{k_{\delta}}\right)<\frac{1}{3}\delta$$ for every $n>n_{\delta}$.

So for $n>n_{\delta}$ we found actually that: $$\mathsf{P}\left(\left|X_{n}Y_{n}-XY\right|>\epsilon\right)\leq\delta$$

Proved is now that $X_{n}Y_{n}\stackrel{p}{\to}XY$ .