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
Best Answer
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 $y$.