I was hoping to type up my proof of Slutsky's Theorem and get confirmation on the excruciating details being all correct...
Statement of Slutsky's Theorem: $$\text{Let }X_n, \ X,\ Y_n,\ Y,\text{ share the same Probability Space }(\Omega,\mathcal{F},P).$$ $$\text{If }Y_n\xrightarrow{\text{prob}}c, \text{ for any constant } c \text{, and }X_n\xrightarrow{\text{dist}}X\text{ then:}$$
$$1.) \ X_n+Y_n\xrightarrow{\text{dist}}X_n+c$$ $$2.) \ X_nY_n\xrightarrow{\text{dist}}cX. \ \ \ \ \ \ \ \ \ \ $$
Proof of 1.)
Let $x$ be a point such that $x-c$ is a point of continuity of $F_x$ and pick $\epsilon$ such that $x-c+\epsilon$ is another point of continuity of $F_x$.
By the Law of Total Probability: $$P(X_n+Y_n\leq x)=P(\{X_n+Y_n\leq x\}\cap\{|Y_n-c|\leq\epsilon\})+P(\{X_n+Y_n\leq x\}\cap\{|Y_n-c|>\epsilon\})$$ However, as $$\{ \ \{X_n+Y_n\leq x\}\cap\{|Y_n-c|>\epsilon\}\ \}\subseteq\{|Y_n-c|>\epsilon\}$$ $$\Rightarrow P(\{X_n+Y_n\leq x\}\cap\{|Y_n-c|>\epsilon\})\leq P(|Y_n-c|>\epsilon)$$ $$\Rightarrow P(X_n+Y_n\leq x)\leq P(\{X_n+Y_n\leq x\}\cap\{|Y_n-c|\leq\epsilon\})+P(|Y_n-c|>\epsilon)$$ Now note that $$\{|Y_n-c|\leq\epsilon\}\equiv\{c-\epsilon \leq Y_n \leq c+\epsilon\}$$ Thus $$P(X_n+Y_n\leq x) \leq P(X_n+c-\epsilon\leq x)+P(|Y_n-c|>\epsilon)$$
This inequality holds because we then have $$X_n+c-\epsilon\leq X_n+Y_n \leq X_n+c+\epsilon$$ so $$P(X_n+Y_n\leq x)\leq P(X_n+c-\epsilon\leq x)$$ and doing our limits $$\displaystyle\limsup_{n}P(X_n+Y_n\leq x) \leq \displaystyle\limsup_{n}P(\{X_n+c-\epsilon\leq x\})+\displaystyle\limsup_{n}P(|Y_n-c|>\epsilon)$$ The far right term is zero by $Y_n\xrightarrow{prob}c$ which gives us $$\displaystyle\limsup_{n}P(X_n+Y_n\leq x) \leq \displaystyle\limsup_{n}P(X_n+c\leq x+\epsilon)$$ As $\epsilon$ is only on the right side, and can be as small as needed and $x-c$ a point of continuity of $F_x$, we get $$\displaystyle\limsup_{n}P(X_n+Y_n\leq x) = P(X+c\leq x)$$ The $\displaystyle\liminf_{n}$ follows similarly giving the same type of result $$\displaystyle\liminf_{n}P(X_n+Y_n\leq x) = P(X+c\leq x)$$ Thus by their equivalence $$\displaystyle\lim_{n\rightarrow\infty}P(X_n+Y_n\leq x) = P(X+c\leq x)\ \ \square$$
Proof of 2.)
Assuming $\pm\epsilon / \delta$ are points of continuity of $F_x$, and WLOG let $Y_n\xrightarrow{prob}0$: $$P(|X_nY_n| > \epsilon)=P(\{|X_nY_n|>\epsilon\}\cap\{|Y_n|>\delta\}) + P(\{|X_nY_n>\epsilon\}\cap\{|Y_n|\leq\delta\})$$ Therefore as $$|Y_n|\leq\delta\Rightarrow\frac{1}{|Y_N|}\geq\frac{1}{\delta}\Rightarrow\frac{|X_nY_n|}{|Y_N|}\geq\frac{\epsilon}{\delta}\Rightarrow|X_n|\geq\frac{\epsilon}{\delta}$$ Also noting $$\{|X_nY_n|>\epsilon\}\cap\{|Y_n|>\delta\}\subseteq\{|Y_n|>\delta\}$$ $$\Rightarrow P(|X_nY_n| > \epsilon) \leq P(|Y_n|>\delta)+P(|X_n|>\frac{\epsilon}{\delta})$$ And so by subadditivity $$\leq P(|Y_n|>\delta) + P(X_n>\frac{\epsilon}{\delta}) + P(X_n\leq\frac{-\epsilon}{\delta}) $$ $$ = P(|Y_n|>\delta) + 1 - P(X_n\leq\frac{\epsilon}{\delta}) + P(X_n\leq\frac{-\epsilon}{\delta})$$ $$=P(|Y_n|>\delta) + 1 - F_n(\frac{\epsilon}{\delta})+F_n(\frac{-\epsilon}{\delta}) $$ Thus $$\limsup_{n}P(|X_nY_n|>\epsilon)\leq\limsup_{n}P(|Y_n|>\delta) + \limsup_{n}1 - \limsup_{n}F_n(\frac{\epsilon}{\delta})+\limsup_{n}F_n(\frac{-\epsilon}{\delta})$$ Using $Y_n\xrightarrow{prob}0$ we simplify to $$\limsup_{n}P(|X_nY_n|>\epsilon)\leq 1 - \limsup_{n}F_n(\frac{\epsilon}{\delta})+\limsup_{n}F_n(\frac{-\epsilon}{\delta})$$ $$=1 - F(\frac{\epsilon}{\delta})+F(\frac{-\epsilon}{\delta})$$ Let $\delta\rightarrow 0$ and so by properties of a CDF $$F(\frac{\epsilon}{\delta})\rightarrow 1 \text{ and }F(\frac{-\epsilon}{\delta})\rightarrow0$$ $$\Rightarrow P(|X_nY_n|>\epsilon)\rightarrow0$$ If $Y_n$ converges to a non zero constant the proof remains by just substitution $Y_n=Z_n+c$. The $\liminf$ again follows similarly.
