Q: My proof of uniqueness of limit (of sequence) Should I try to write a direct (i.e. non$-$by-contradiction) proof
  instead of the below proof?
  (I was told that mathematicians prefer direct proofs.)
We consider a convergent sequence
  which we denote by
  $(x_n)_{n \in \mathbb{N}}$.
  By definition, there is a limit (of the sequence).
$\textbf{Theorem.}$
  There are no two limits.
$\textit{Proof.}$
  We prove by contradiction.
  To that end,
  we assume that there are two limits.
  Now, our mission is to deduce a contradiction.
  Let $x,x'$ be limits such that $x \ne x'$.
  By definition ($\textit{limit}$), we have
  \begin{equation*}
    \begin{split}
     &\forall \varepsilon \in \mathbb{R}, \varepsilon > 0 :
      \exists N           \in \mathbb{N}                  :
      \forall n           \in \mathbb{N}, n > N           :
      |x_n - x| < \varepsilon && \text{ and} \\
     &\forall \varepsilon \in \mathbb{R}, \varepsilon > 0 :
      \exists N           \in \mathbb{N}                  :
      \forall n           \in \mathbb{N}, n > N           :
      |x_n - x'| < \varepsilon.
    \end{split}
  \end{equation*}
Since $x \ne x'$, we have $0 < \frac{1}{2} |x - x'|$.
  We choose $\varepsilon := \frac{1}{2} |x - x'|$.
By assumption, there are $N,N' \in \mathbb{N}$ such that
  \begin{equation*}
    \begin{split}
     &\forall n           \in \mathbb{N}, n > N           :
      |x_n - x| < \varepsilon && \text{ and} \\
     &\forall n           \in \mathbb{N}, n > N'          :
      |x_n - x'| < \varepsilon.
    \end{split}
  \end{equation*}
  We choose $n := \max\{N, N'\} + 1$.
  Obviously, both $n > N$ and $n > N'$.
  Therefore, we have both $|x_n - x| < \varepsilon$ and $|x_n - x'| < \varepsilon$.
  Thus, by adding inequalities,
  \begin{equation*}
    |x_n - x| + |x_n - x'| < 2 \varepsilon .
  \end{equation*}
  Moreover,
  \begin{equation*}
    \begin{split}
      2 \varepsilon & =   |x - x'|
                      && | \text{ by choice of } \varepsilon \\
                    & =   |x + 0 - x'| \\
                    & =   |x + ( - x_n + x_n) - x'| \\
                    & =   |(x - x_n) + (x_n - x')| & \qquad & \\
                    & \le |x - x_n| + |x_n - x'|
                      && | \text{ by subadditivity of abs. val.} \\
                    & =   |x_n - x| + |x_n - x'|
                      && | \text{ by evenness of abs. val.} \\
    \end{split}
  \end{equation*}
  Hence, by transitivity, we have $2 \varepsilon < 2 \varepsilon$.
  Obviously, we deduced a contradiction. QED

A: For a slightly more abstract answer, recall that every metric space is a Hausdorff space, i.e. distinct points can be separated by neighborhoods (you can easily prove this as $d(x,y)>0$ if $x\ne y$).
So if $x_n\to x$ and $y\ne x$, let $U$ and $V$ be open balls centered at $x$ and $y$, respectively, such that $U\cap V=\varnothing$. Then there exists $N$ such that $n\geqslant N$ implies $x_n\in U$, and hence $x_n\notin V$, so that $x_n$ cannot converge to $y$.

A: There is nothing wrong with a proof by contradiction, and your proof appears correct. You can prove it directly though if you wish. If you let $\varepsilon>0$ be given, then we know that there exists $N, M\in\mathbb{N}$ such that
\begin{align*}
|x_n-x|<\frac{\varepsilon}{2}\ \hspace{10pt}\text{for all $n>N$} \\
\text{and}\ |x_n-x'|<\frac{\varepsilon}{2} \hspace{10pt}\text{for all $n>M$}.
\end{align*}
Now set $N'=\max\{N, M\}$, then we see that for all $n>N'$ 
\begin{align*}
|x-x'|&=|x-x_n+x_n-x'| \\
&\leq |x-x_n|+|x_n-x'| \\
&<\varepsilon.
\end{align*}
This is true for all $\varepsilon>0$, and so we can take this value as small as we like. Therefore $|x-x'|=0$ from which $x-x'=0$ and $x=x'$. 
A little note about notation though, the colon $:$ means 'such that'. This part of your proof 
$$ \forall\varepsilon\in\mathbb{R}, \varepsilon>0:\exists N\in\mathbb{N}:\forall n\in\mathbb{N}, n>N :|x_n-x|<\varepsilon $$
reads 'For all $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$ such that there exists $N\in\mathbb{N}$ such that for all $n>N$ such that $|x_n-x|<\varepsilon$', which doesn't really make sense. The correct statement is 'For all $\varepsilon\in\mathbb{R}$ with $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that for all $n>N$ we have $|x_n-x|<\varepsilon$' and you can write this
$$ \forall\varepsilon\in\mathbb{R}, \varepsilon>0, \exists N\in\mathbb{N} : \forall n>N\ \text{we have}\ |x_n-x|<\varepsilon.
$$

A: You can prove it in a simpler way. Suppose $(x_n)_{n\in\mathbb{N}}$ has a limit $L$, if $M\neq L$ is another limit for the sequence, this will lead to a contradiction.
We can write $M = L+\delta$, for some $\delta\neq0$, now let $0<\varepsilon < |\delta|$. There is a $n_0\in\mathbb{N}$ such that $n >n_0 \implies |x_n-L| < \varepsilon/2$. 
If $|x_n-M| < \varepsilon/2$ for some $n>n_0$ (this has to happen if $M$ is also a limit for the sequence), then $|M-L| = |M-x_n-(L-x_n)|\leq|M-x_n| + |L-x_n| < \varepsilon < |\delta|$. Which is a contradiction, because $|M-L| = |\delta|$.
