Show that the unit sphere is a complete metric space equipped with $d(x,y):= \arccos \langle x,y \rangle_{\mathbb{R}^n}$. So we were shown this problem from our first functional analysis lecture and I was wondering if someone could help or give a hint:

Show that the unit sphere $\mathbb{S}^{n-1} := \{ x \in \mathbb{R}^n: \| x \| = 1 \}$ is a complete metric space equipped with $d(x,y):= \arccos \langle x,y \rangle_{\mathbb{R}^n}$ where $\langle  x,y \rangle_{\mathbb{R}^n}$ denotes the standard dot product. 

We're having no trouble showing positivity and symmetry but could use help showing completeness and the triangle inequality.
Kind regards
Edit:
So with MatiasHeikkilä's help, what's left to show is that $\theta_{x,y} \leq \theta_{x,z} + \theta_{y,z}$ if I'm not mistaken. Do I need to make a case-by-case proof to show this inequality or is there an easier way? I would argue something  along the lines of: "$z$ lies between $x$ and $y$ implies the equality", "$z$ does not lie on the (smallest) path between $x$ and $y$ but $z$ lies on the half-sphere with $x$ and $y$ on it implies the inequality (since $\theta_{x,y} < \theta_{x,z}$) and lastly $z$ does neither lie on the (smallest) path between $x$ and $y$ nor does $z$ lie on the half-sphere implies the inequality (since $\theta_{x,z} + \theta_{y,z} = 2 \pi - \theta_{x,y} \geq \pi $)
 A: Let us first show the triangle inequality for $d$, i.e. $d\left(x,z\right)\leq d\left(x,y\right)+d\left(y,z\right)$.
Since we have $\left\langle Ux,Uy\right\rangle =\left\langle x,y\right\rangle $
for every orthogonal matrix, by choosing $U$ with $Uy=e_{1}=\left(1,0,\dots,0\right)$,
we can assume $y=e_{1}$ (in short, we used that $d$ is invariant
under orthogonal transformations). Since the cosine is strictly decreasing
on $\left[0,\pi\right]$ (and since $\arccos$ takes values in $\left[0,\pi\right]$),
the desired inequality is equivalent to
\begin{align*}
\left\langle x,z\right\rangle  & \overset{!}{\geq}\cos\left(\arccos\left\langle x,e_{1}\right\rangle +\arccos\left\langle e_{1},z\right\rangle \right)\\
 & =\cos\left(\arccos x_{1}+\arccos z_{1}\right)\\
\left(\text{by trigonometric formulas}\right) & =\cos\left(\arccos x_{1}\right)\cos\left(\arccos z_{1}\right)-\sin\left(\arccos x_{1}\right)\sin\left(\arccos z_{1}\right)\\
\left(\text{since }\sin\geq0\text{ and hence }\sin=\sqrt{1-\cos^{2}}\text{ on }\left[0,\pi\right]\right) & =x_{1}z_{1}-\sqrt{1-\cos^{2}\left(\arccos x_{1}\right)}\sqrt{1-\cos^{2}\left(\arccos z_{1}\right)}\\
 & =x_{1}z_{1}-\sqrt{1-x_{1}^{2}}\sqrt{1-z_{1}^{2}}.
\end{align*}
By rearranging, we see that this inequality is equivalent to
\begin{align*}
-\sum_{i=2}^{n}x_{i}z_{i} & =x_{1}z_{1}-\left\langle x,z\right\rangle \\
 & \overset{!}{\leq}\sqrt{1-x_{1}^{2}}\sqrt{1-z_{1}^{2}}\\
\left(\text{since }1=\left|x\right|^{2}=\sum_{i=1}^{n}x_{i}^{2}\right) & =\sqrt{\sum_{i=2}^{n}x_{i}^{2}}\sqrt{\sum_{i=2}^{n}z_{i}^{2}}.
\end{align*}
But by Cauchy Schwarz, we have
$$
-\sum_{i=2}^{n}x_{i}z_{i}\leq\left|\sum_{i=2}^{n}x_{i}z_{i}\right|\leq\sqrt{\sum_{i=2}^{n}x_{i}^{2}}\sqrt{\sum_{i=2}^{n}z_{i}^{2}},
$$
so that the inequality above holds. We have thus verified the triangle
inequality. By what you have already shown, $d$ is a metric.
For completeness, I leave it to you to verify that
$$
\Phi:\left(S^{n-1},\mathcal{T}\right)\to\left(S^{n-1},d\right),x\mapsto x
$$
is continuous, where $\mathcal{T}$ denotes the usual topology. Thus,
$\left(S^{n-1},d\right)$ is compact. Now, it is well known that every
compact metric space is complete, so that we are done.
A: For $x$, $y\in S^{n-1}$ write $\arccos\langle x,y\rangle=:\theta_{xy}$. Then $$d(x,y):=\theta_{xy}\in[0,\pi]$$ is nothing else but the usual spherical distance on $S^{n-1}$. Using elementary geometry one immediately verifies that 
$$|x-y|=2\sin{\theta_{xy}\over 2}\qquad(x, \>y\in S^{n-1})\ .\tag{1}$$
In particular, $\theta_{xy}$ is a strictly increasing function of the eucliden distance $|x-y|$.
In order to prove the triangle inequality it is sufficient to consider  three points $x$, $y$, $z\in S^2$. If $\theta_{xy}+\theta_{yz}\geq\pi$ we immediately obtain $$d(x,z)\leq\pi\leq d(x,y)+d(y,z)\ .$$ Therefore assume $\theta_{xy}+\theta_{yz}<\pi$. Consider $y$ as north pole, and rotate the point $z$ around the axis $Oy$ to a position $z'$ such that $z'$ and $x$ are on opposite meridians. We then have
$$|x-z|\leq |x-z'|$$
and therefore
$$d(x,z)\leq d(x,z')=d(x,y)+d(y,z')=d(x,y)+d(y,z)\ .$$
The completeness of $(S^{n-1},d)$ is seen as follows: From $${2\over \pi}\>t\leq \sin t\leq t\qquad\left(0\leq t\leq{\pi\over2}\right)$$
and $(1)$ one concludes that 
$$|x-y|\leq d(x,y)\leq{\pi\over2}\>|x-y|\qquad(x, \>y\in S^{n-1})\ .$$
This immediately implies that the euclidean metric $|\cdot|$ and $d$ are equivalent: A sequence on $S^{n-1}$ that converges with respect to $|\cdot|$ converges with respect to $d$ as well, and vice versa. 
But more than that: A sequence that is Cauchy with respect to $|\cdot|$ is Cauchy as well with respect to $d$, and vice versa. This allows to conclude that the metric space $(S^{n-1},d)$ is indeed complete.
