You have $$ \begin{aligned} &\sup_{n\ge1}\sqrt{Ed\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2} \\ &=\sup_{n\ge1}\|d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) \|_2 \leq 2 \sqrt{E \left[ X_1^2; |X_1| > v \right] } \text{ for all } v>0. \end{aligned} $$ So, by the Chebyshev/Markov inequality, $$ \begin{aligned} &\lim_{v \to \infty} \sup_n P \left( d \left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) > \lambda \right) \\ &\le\lim_{v \to \infty} \sup_n \frac{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]}{\lambda^2} \\ &\le\lim_{v \to \infty} \frac{4E \left[ X_1^2; |X_1| > v \right] } {\lambda^2} =0, \end{aligned} $$ because, by the dominated convergence theorem, $E \left[ X_1^2; |X_1| > v \right]\to0$ as $v\to\infty$.

You have $$ \begin{aligned} &\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]} \\ &=\sup_{n\ge1}\|d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) \|_2 \leq 2 \sqrt{E \left[ X_1^2; |X_1| > v \right] } \text{ for all } v>0. \end{aligned} $$ So, by the Chebyshev/Markov inequality, $$ \begin{aligned} &\lim_{v \to \infty} \sup_n P \left( d \left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) > \lambda \right) \\ &\le\lim_{v \to \infty} \sup_n \frac{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]}{\lambda^2} \\ &\le\lim_{v \to \infty} \frac{4E \left[ X_1^2; |X_1| > v \right] } {\lambda^2} =0, \end{aligned} $$ because, by the dominated convergence theorem, $E \left[ X_1^2; |X_1| > v \right]\to0$ as $v\to\infty$.