Show for $X_0,X_1,...$ RV that $X_n\to^p X_0$ iff $\max_{k=1,...,d}\vert X_{n,k}-X_{0,k}\vert \to^p 0$ iff $\sum_{k=1}^d(X_{n,k}-X_{0,k})^2\to^p 0$ 
Show for $X_0,X_1,...$ $\mathbb{R}^d$ random vectors on the same probability space, that $X_n\to^p X_0$ if and only if $\max_{k=1,...,d}\vert X_{n,k}-X_{0,k}\vert \to^p 0$ if and only if $\sum_{k=1}^d(X_{n,k}-X_{0,k})^2\to^p 0$

$\vert X_{n,k}-X_{0,k}\vert$ is the kth coordinate distances of $X_n$ and $X_0$
I have this continuous mapping theorem which says if $g$ is a continuous real function and $X_n\to^p X$ then $g(X_n)\to^p g(X)$.
I want to prove them as $1\to 2\to 3\to 1$:
1. Assume $X_n\to^p X_0$
Let $\epsilon >0$
Then $P(\vert X_n-X_0\vert >\epsilon)\to 0$ as $n\to \infty$
I assumed $\vert X_n-X_0\vert$ is the same as euclidean distance, in which case I can use that $\vert X_{n,k} -X_{0,k}\vert\leq \vert X_n-X_0\vert$ for any $k=1,...,d$
And thus $\max_{k=1,...,d}\vert X_{n,k}-X_{0,k}\vert\leq \vert X_n-X_0\vert$
Then $P(\vert X_n-X_0\vert>\epsilon)\geq P(\max_{k=1,..,d} \vert X_{n,k}-X_{0,k}\vert>\epsilon)$
Then by squeeze theorem since $P(\vert X_n-X_0\vert>\epsilon)\to 0$, $P(\max_{k=1,..,d} \vert X_{n,k}-X_{0,k}\vert>\epsilon)\to 0$ also.
2. Assume $\max_{k=1,...,d}\vert X_{n,k}-X_{0,k}\vert \to^p 0$
Let $\epsilon>0$
Let $Y_n=\max_{k=1,...,d}\vert X_{n,k}-X_{0,k}\vert$
$\sum_{k=1}^d (X_{n,k}-X_{0,k})^2 \leq dY_n^2$
By continuous mapping, we know that since $Y_n\to^p 0$ then so does $dY_n^2\to^p 0$
And since $\sum_{k=1}^d (X_{n,k}-X_{0,k})^2 \leq dY_n^2$, $\lim_{n\to \infty} P(\vert \sum_{k=1}^d (X_{n,k}-X_{0,k})^2\vert>\epsilon)\leq \lim_{n\to \infty}P(dY_n^2>\epsilon)=0$ so by squeeze theorem  $P(\vert \sum_{k=1}^d (X_{n,k}-X_{0,k})^2\vert>\epsilon)=0$, as required.
3. Assume $\sum_{k=1}^d(X_{n,k}-X_{0,k})^2\to^p 0$
Since $\vert X_n-X_0\vert = \sqrt{\sum_{k=1}^d(X_{n,k}-X_{0,k})^2}$
Since square root is continuous by continuous mapping theorem we get $\vert X_n - X_0\vert\to^p 0$
And since absolute value is continuous, we get $X_n-X_0\to^p 0$
Then $\lim_{n\to \infty} P(\vert X_n-X\vert>\epsilon =\lim_{n\to \infty} P(\vert X_n-X-0\vert>\epsilon=0$ so $X_n\to^p X$
Is this correct?
I think most of this is correct assuming I can treat distances between random vectors the same as regular vectors in $\mathbb{R}^n$, so I'm wondering if that is true. And also whether I did something in this proof which is not actually true for random variables.
 A: This looks alright to me (algebraic identities can indeed be applied to random vectors), except I'm not sure I follow your step from 3 to 1 as currently written; the 2-norm is a continuous map from $\mathbb R^n$ to $\mathbb R$ but doesn't have an inverse map so I don't see how you would use the continuous mapping theorem (CMT) to move from $\|X_n-X\|_2\overset{p}{\rightarrow}0$ to $X_n\overset{p}{\rightarrow}X,$ where $\|\cdot\|_2$ is the 2-norm.
Fortunately, this step is salvageable by simply noting that
$$P(|\|X_n-X\|_2-0|>\epsilon)=P(\|X_n-X\|_2>\epsilon),$$
so it is clear that $\|X_n-X\|_2\overset{p}{\rightarrow}0\iff X_n\overset{p}{\rightarrow}X.$

I would like to point out that you can show a more general result than the ones you show, namely that $X_n\overset{p}{\rightarrow}X$ if and only if $\|X_n-X\|_p \overset{p}{\rightarrow}0$ for any $p$-norm $\|\cdot \|_p,p\geq 1.$ You essentially proved this for the cases $p=2$ and $p=\infty.$
You can prove this broader claim by showing that for any $p$-norm,
$$X_{n,k} \overset{p}{\rightarrow}X_k,\forall k\in\{1,...,d\}\iff \|X_n-X\|_p \overset{p}{\rightarrow}0.$$
