real analysis - 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