Proof regarding the limit of arctanx. 
So we're very limited in what we can and cannot use regarding properties of trigonometric functions since we haven't proven much, so I was forced to proof it the long way which invites room for mistakes. I was wondering if you have any tips regarding shortening the proof or if something in the proof doesn't add up..
Consider the function $ f: R \to R \ $, $f(x) = \arctan(x).$
First we show the limits of $f$ exist.
Claim: $\lim _{x\to \infty }\left(\arctan\left(x\right)\right)=\frac{\pi }{2}$
Proof: Let $ϵ > 0$.
We choose $M = \max \left\{\tan\left(\frac{\pi}{2}-ϵ\right),1\right\}>0$. Then for every $x>M$ we have $|\arctan(x)-\frac{\pi}{2}|<ϵ$
Note that we've seen in class the function tan is strictly increasing for every $x ∈ (-\frac{\pi}{2},\frac{\pi}{2})$,
and since arctan is the inverse of tan, it's also strictly increasing.
Therefore ★$\arctan(x) > \arctan(y) \implies -\arctan(x) < -\arctan(y)$ for every $x > y$.
Also note we've seen in class that $-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$ for every $x ∈ R \ $, therefore $\arctan(x) - \frac{\pi}{2} < 0$.
$$\begin{align}|\arctan(x)-\frac{\pi}{2}|&=\frac{\pi}{2}-\arctan\left(x\right)\\
&\frac{★}{<} \ \frac{\pi}{2}-\arctan\left(M\right)\\
&=\frac{\pi}{2}-\arctan\left(\tan\left(\frac{\pi}{2}-ϵ\right)\right)\\
&=\frac{\pi}{2}-\frac{\pi}{2}+ϵ\\
&≤ϵ\end{align}$$
Therefore the limit holds.
Claim: $\lim _{x\to -\infty }\left(\arctan\left(x\right)\right)=-\frac{\pi }{2}$
Proof: Let $ϵ > 0$.
We choose $M = max \left\{\tan\left(-\frac{\pi}{2}+ϵ\right),-1\right\}>0$. Then for every $x<M$ we have $|\arctan(x)+{\pi}{2}|<ϵ$
Note we've seen in class that $-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$ for every $x ∈ R \ $, therefore $\arctan(x) + \frac{\pi}{2} > 0$.
$$\begin{align}|\arctan(x)-\frac{\pi}{2}|&=\frac{\pi}{2}+\arctan\left(x\right)\\
&\frac{★}{<} \ \frac{\pi}{2}+\arctan\left(M\right)\\
&=\frac{\pi}{2}+\arctan\left(\tan\left(-\frac{\pi}{2}+ϵ\right)\right)\\
&=\frac{\pi}{2}-\frac{\pi}{2}+ϵ\\
&≤ϵ\end{align}$$
Therefore the limit holds.
Now we show that $f$ has no extreme points.
Note that $f$ is strictly increasing, therefore
for every $x,y ∈ R$ such that $x < y$, we have $f(x) < f(y)$.
Assume toward contradiction that $f$ has a maximum, we will denote it $a$.
Since f is strictly increasing, there exist a point $b ∈ R$ such that $a < b$ for which $f(a) < f(b)$,
in contradiction to the minimality of $a$.
Assume toward contradiction that $f$ has a minimum, we will denote it $c$.
Since f is strictly increasing, there exist a point $d ∈ R$ such that $d < c$ for which $f(d) < f(c)$,
in contradiction to the minimality of $c$.
 A: 
Consider the function $ f: R \to R \ $, $f(x) = \arctan(x).$

A much easier choice would have been something like
$$f(x) = \begin{cases}-\dfrac 1x,&\quad x\le -3\\
\dfrac 12 +\dfrac x{18},&\quad-3 < x < 3\\
1 - \dfrac 1x,&\quad 3\le x\end{cases}$$
Continuity is easy to establish (particularly if you already have the theorem about piecing together continuous functions that agree at the seam giving a continuous result). And it is also simple to show that it is strictly increasing. Showing the limits are $0$ and $1$ is also much simpler than yours.
Perhaps even easier would be $$f(x) = \dfrac x{|x| + 1}$$
Overall, your proof is good. Almost none of the things I bring up are logical errors, and the exceptions are trivial technical issues. But they do distract from your argument.

Claim: $\lim _{x\to \infty }\left(\arctan\left(x\right)\right)=\frac{\pi }{2}$
Proof: Let $ϵ > 0$.

You need to restrict $0 < \epsilon < \pi$ here. Otherwise some of your later claims become false. This is not an issue for the definition of a limit, because if something is $<$ this restricted $\epsilon$, it is also $<$ all larger values of $\epsilon$, which is all that is required for them.

We choose $M = \max \left\{\tan\left(\frac{\pi}{2}-ϵ\right),1\right\}>0$.

You make no use of the restriction that $M \ge 1$. In fact, later on, you ignore the possibility that $M$ could be $1$ instead of $\tan\left(\frac{\pi}{2}-ϵ\right)$. You should just set $M = \tan\left(\frac{\pi}{2}-ϵ\right)$ instead.

Then for every $x>M$ we have $|\arctan(x)-\frac{\pi}{2}|<ϵ$
Note that we've seen in class the function tan is strictly increasing for every $x ∈ (-\frac{\pi}{2},\frac{\pi}{2})$, and since arctan is the inverse of tan, it's also strictly increasing.

Note that you've already made use once of the increasing behavior of $\tan$ in saying that for $x > M, \left|\arctan x - \frac \pi2\right| < \epsilon$

Therefore ★$\arctan(x) > \arctan(y) \implies -\arctan(x) < -\arctan(y)$ for every $x > y$.

This is convoluted, and all you really need from it is that if $x > M$, then $\arctan x > \arctan M$, and therefore $\frac \pi2 - \arctan x < \frac \pi2 - \arctan M$. I suggest explaining that directly.

Also note we've seen in class that $-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$ for every $x ∈ R \ $, therefore $\arctan(x) - \frac{\pi}{2} < 0$.

You need to insert "Let $x > M$" here, as your last reference to $x$ is as an arbitrary element of $\Bbb R$. But you are about to do something that requires $x > M$.

$$\begin{align}|\arctan(x)-\frac{\pi}{2}|&=\frac{\pi}{2}-\arctan\left(x\right)\\
&\frac{★}{<} \ \frac{\pi}{2}-\arctan\left(M\right)\\
&=\frac{\pi}{2}-\arctan\left(\tan\left(\frac{\pi}{2}-ϵ\right)\right)\\
&=\frac{\pi}{2}-\frac{\pi}{2}+ϵ\\ &≤ϵ\end{align}$$
Therefore the limit holds.

Other than the weird symbol where you apparently mean $\le$, and forgetting that $M$ could be $1$, this is good.

Claim: $\lim _{x\to -\infty }\left(\arctan\left(x\right)\right)=-\frac{\pi }{2}$

The whole repeat proof is unnecessary. I don't know which of the many possible developments of trigonometric functions your course follows, but in all of them the fact that $\sin$ is odd and $\cos$ is even is almost trivial. From this it follows that $\tan$ is odd and therefore (with proper choice of domain) so is $\arctan$. Hence $$\lim_{x \to -\infty} \arctan x = \lim_{t \to \infty} \arctan(-t) = -\lim_{t\to\infty} \arctan t = -\frac\pi 2$$

Now we show that $f$ has no extreme points.
Note that $f$ is strictly increasing, therefore
for every $x,y ∈ R$ such that $x < y$, we have $f(x) < f(y)$.
Assume toward contradiction that $f$ has a maximum, we will denote it
$a$.

The way this is stated, $a$ would be the maximum value that $f$ takes on, but what you mean is "$f$ takes on a maximum at $a$", so it is $f(a)$ which is the maximum, not $a$.

Since f is strictly increasing, there exist a point $b ∈ R$ such that $a < b$ for which $f(a) < f(b)$,
in contradiction to the minimality [maximality] of $a$.

Other than the wording errors, this argument and the minimum are correct, but overly verbose. Simply noting that for any $x$, if $y < x$ then $f(y) < f(x)$ and if $y > x$ then $f(y) > f(x)$ is sufficient to recognize that $x$ cannot be a maximum or minimum, even locally (because there are always $y$ above or below $x$ - if the domain were closed and bounded instead, then $f$ would have extrema at the endpoints).
