If $X_n\xrightarrow[]{p}c$, does this imply that $\vert X_n\vert\xrightarrow[]{p}\vert c\vert$?
Since $X_n\xrightarrow[]{p}c$, we know that $\mathbb{P}(\vert X_n - c\vert>e)\rightarrow 0$. By the reverse triangle inequality, we know $\big \vert X_n - c\big\vert \geq\big\vert \vert X_n \vert- \vert c\vert \big\vert$, which implies that $\{w : \big\vert \vert X_n \vert- \vert c\vert \big\vert>\epsilon\}\subset\{w : \big\vert X_n - c \big\vert>\epsilon\}$. This shows that $\mathbb{P}(\big\vert \vert X_n \vert- \vert c\vert \big\vert>\epsilon)\leq\mathbb{P}(\big\vert X_n - c \big\vert>\epsilon)\rightarrow 0$, which proves the claim?
Yes your proof is correct. The same argument shows that if $f$ is a Lipschitz continuous function and $X_n\to X$ in probability, then $f\left(X_n\right)\to f(X)$ in probability. Actually, it also work for continuous functions, but the proof is a little bit more technical.