real analysis - Proving part of the theorem that a continuous function on the bounded interval is uniformly continuous

I am asked to prove the following. Suppose $D$ is a closed and bounded subset of $R$ and suppose $f:D\to R$ is continuous on $D$. Then $f$ is uniformly continuous:
Proof: Suppose by contradiction that $\{x_n\}$ and $\{y_n\}$ are sequences in $D$ s.t.

$$\lim_{n\to\infty} [x_n-y_n] =0$$

and the sequence $[f(x_n)-f(y_n)]$ does not converge to $0$.

We want to negate the statement of convergence for the above s.t. we have: "There exists an $\epsilon>0$ s.t. for all natural number k there exists an $n_k>k$ s.t. |$f(x_{n_k})-f(v_{n_k})$|$\geq \epsilon$.

Now apply the sequential compactness theorem to $\{x_{n_k}\}$ which is a subset of $D$. There is a limit point $x_0$ s.t. there is a further subsequence $\{x_{n_{k_j}}\}$ s.t. the limit as j tends to infinity is $x_0$, which is in $D$.

Now here's what I am supposed to prove: By supposing that $\{x_n\}$ and $\{y_n\}$ are sequences satisfying the fact that the image of a continuous function on a closed bounded interval is bounded above and assuming $\{x_n\}$ converges to $x$ I want to show that the limit as $j$ tends to infinity of $y_{n_{k_j}}$ is $x_0$. How would you do this part?

So basically I am just proving this one last part of the proof.