Is this proof of continuity of every metric circular? Say I want to show every metric defined on $X$ is continuous, where $X$ is an arbitrary metric space with the metric $d$. That is I want to show, that for two convergent sequences $x_n\rightarrow x$ and $y_n \rightarrow y$ it holds, that :
$$ \lim\limits_{n\rightarrow \infty}d(x_n,y_n)=d(x,y)$$

Proof: With some algebraic manipulation of the triangle inequality we
  find that for arbitrary values $a,b,c,e \in X$ it holds that  $$
 |d(a,b)-d(c,e)|\leq d(a,c)+d(e,b)$$
and then it follows that 
$$ |d(x_n,y_n)-d(x,y)| \leq d(x_n,x)+d(y_n,y) $$
Now if we take the limit as $n \rightarrow \infty$ it follows from the
  convergence of $x_n$ and $y_n$ that $ |d(x_n,y_n)-d(x,y)| \leq 0 $,
  hence the result.

But in the last step, we are using that $\lim\limits_{n\rightarrow \infty}d(x_n,x) = d(x,x)=0$, i.e. we are implicitly using that the metric is a continuous mapping, by "pulling" the limit inside of the function. 
Does this make the proof invalid? If yes, how can one avoid circular reasoning here?
 A: Given that $(X,d)$ is a metric space. If $(x_n)_{n \in \mathbb{N}}$ is a sequence in $X$ and $x \in X$, then the statement $``x_n \to x$ as $n \to \infty"$ is defined as $``d(x_n,x) \to 0$ as $n \to \infty"$, where the latter is with respect to standard continuity of the reals.
In other words, we have the following equivalence:


*

*$x_n \to x$ as $n \to \infty$

*$\lim_{n \to \infty} d(x_n,x) = 0$



Now as you pointed out, we have the inequality $|d(x_n,y_n) - d(x,y)| \leq d(x_n,x) + d(y_n,y)$. You can think of the left and right hand side as real-valued sequences, which we know how to deal with.
Limits of real-valued sequences preserve strict inequalities and sums, therefore:
$$\lim_{n \to \infty} |d(x_n,y_n) - d(x,y)| \leq \lim_{n \to \infty} [d(x_n,x) + d(y_n,y)]$$
$$\lim_{n \to \infty} |d(x_n,y_n) - d(x,y)| \leq \lim_{n \to \infty} d(x_n,x) + \lim_{n \to \infty} d(y_n,y)$$
Therefore $\lim_{n \to \infty} |d(x_n,y_n) - d(x,y)| \leq 0$, by definition of convergence.

However to get right to the point of your question, you are not using circular reasoning to use $\lim_{n \to \infty} d(x_n,x) = 0$, but not because $\lim_{n \to \infty} d(x_n,x) = d(\lim_{n \to \infty} x_n,x)$. We have $\lim_{n \to \infty} d(x_n,x) = 0$ because that is the definition of $(x_n)_{n \in \mathbb{N}}$ converging to $x$.
A: Please note: Continuity in metric spaces defined via a metric is equivalent to the topological definition of continuity through open sets. If that's what you are seeking you can find a proof here.
But given your description I believe you simply want to prove the statement: "for two convergent sequences $x_n\rightarrow x$ and $y_n \rightarrow y$ it holds, that :
$ \lim\limits_{n\rightarrow \infty}d(x_n,y_n)=d(x,y)$".
HINT:
Given some $\epsilon > 0$ use the properties of $d$, and the fact that $y_n \rightarrow y$ to show that there is some $N_1,N_2 \in \mathbb{N}$ such that for any $z \in X$: $$n>N_1 \Rightarrow |d(y_n,z) - d(y,z)|<\frac{\epsilon}{6} \\ n>N_2 \Rightarrow |d(x_n,z) - d(x,z)|<\frac{\epsilon}{6}$$
Next, use the traingle inaequlity and to show that $n>\max\{N_1,N_2\} \Rightarrow |d(x_n,y_n) - d(x,y)| < \epsilon$.
By definition, it implies that $d(x_n,y_n)$, as a sequence in $\mathbb{R}$, converges to $d(x,y)$.
