Continuous Mapping Theorem (for convergence in probability), Help in understanding proof The following theorem and proof are given in my econometrics lecture notes.
Theorem:
Let $x_{n}, x$ be $k \times 1$ vectors. Assume that $x_{n} \stackrel{P}{\longrightarrow} x,$ and let $g$ be a continuous function in the domain of definition of $x.$ Then $g\left(x_{n}\right) \stackrel{P}{\longrightarrow} g(x)$
Proof:
For any arbitrary small $\varepsilon>0,$ we can choose a compact set $S$ such that
\begin{equation*}
\operatorname{Pr}\{x \notin S\} \leq \frac{1}{2} \varepsilon
\end{equation*}
Because $g$ is continuous, it implies that $g$ is uniformly continuous in $S$ . Hence, for any given $\eta>0,$ we have that $\exists \delta(\eta)>0$ (independent of $x \in S)$ such that
$$ \|x-y\| \leq \delta \Longrightarrow\|g(x)-g(y)\| \leq \eta $$
Because $x_{n} \stackrel{P}{\longrightarrow} x,$ we can choose $n_{0}(\varepsilon, \delta)>0$ such that $\forall n \geq n_{0}$
$$\operatorname{Pr}\left\{\left\|x_{n}-x\right\|>\delta\right\} \leq \frac{1}{2} \varepsilon$$
Alternatively,
$$
\begin{aligned} 
1-\frac{1}{2} \varepsilon & \leq \operatorname{Pr}\left\{\left\|x_{n}-x\right\| \leq \delta\right\} \\ &=\operatorname{Pr}\left\{\left\|x_{n}-x\right\| \leq \delta, x \in S\right\} +\operatorname{Pr}\left\{\left\|x_{n}-x\right\| \leq \delta, x \notin S\right\} \\ & \leq \operatorname{Pr}\left\{\left\|x_{n}-x\right\| \leq \delta, x \in S\right\}+\frac{1}{2} \varepsilon \\ 
\text { by def. of continuity } & \leq \operatorname{Pr}\left\{\left\|g\left(x_{n}\right)-g(x)\right\| \leq \eta, x \in S\right\}+\frac{1}{2} \varepsilon \\ 
& \leq \operatorname{Pr}\left\{\left\|g\left(x_{n}\right)-g(x)\right\| \leq \eta\right\}+\frac{1}{2} \varepsilon 
\end{aligned}
$$
Thus:
$$1-\varepsilon \leq \operatorname{Pr}\left\{\left\|g\left(x_{n}\right)-g(x)\right\| \leq \eta\right\}$$
and the result follows since $\eta$ and $\varepsilon$ are arbitrary positive constants. $\square$
Firstly I assume that the "domain of definition of $x$" is just the codomain of $x \subseteq \mathbb{R}^k$. 
My first question in the proof is why we bother to partition into compact and non-compact sets. Would it not be correct to just proof directly by:
Fix $\varepsilon>0 .$ Continuity of $g(x_n)$ at $x$ means that there exists $\delta>0$ such that $\|x_n-x\| \leq$
$\delta$ implies $\|g(x_n)-g(x)\| \leq \varepsilon .$ Then 
$$\mathbb{P}\left(\left\|g\left(x_{n}\right)-g(x)\right\| \leq \varepsilon\right) \geq \mathbb{P}\left(\left\|x_{n}-x\right\| \leq \delta\right) \rightarrow 1$$
Secondly, for the original proof how can we be guaranteed to find a compact set $S$ such that $\operatorname{Pr}\{x \notin S\} \leq \frac{1}{2} \varepsilon$?
 A: 
My first question in the proof is why we bother to partition into compact and non-compact sets.

Continuity of $g$ at $x$ gives you a number $\delta$ which is dependent on the value of $x$. You would thus have to define a (measurable) function $\delta(x)$ such that $\Vert x_{n} - x \Vert \leq \delta(x)$ implies $\Vert g(x_{n}) - g(x) \Vert \leq \varepsilon$ and then consider
$$
\mathbb{P}\left(\Vert x_{n} - x \Vert \leq \delta(x) \right)
$$
However, the fact the $x_{n}$ converges to $x$ in probability does not allow you to conclude that this probability goes to one. In order to appeal to that definition, you must provide a fixed real number $\delta$, not a random variable $\delta(x)$. 

Secondly, for the original proof how can we be guaranteed to find a compact set $S$ such that $\Pr \lbrace x\notin S\rbrace \leq \frac{1}{2}\varepsilon$?

Here we can take the sequence of rectangles $\left\lbrace[-n, n]^{k}\right\rbrace_{n\in\mathbb{N}}$ in $\mathbb{R}^{k}$. This is a countable, increasing sequence whose union is $\mathbb{R}^{k}$. The bound on the probability then follows from continuity of measure (see for example https://en.wikipedia.org/wiki/Measure_%28mathematics%29#Continuity_from_below).
