I'm attempting to prove the following statement:
"Suppose that $f:X \to Y$ is a uniformly continuous function between metric spaces. Suppose that $\{x_n\}$ and $\{x'_n\}$ are two sequences in $X$ such that $d_X(x_n,x'_n) \to 0$ in $\mathbb{R}$. Show that $d_Y(f(x_n),f(x'_n)) \to 0$.
Since $d_X(x_n,x'_n) \to 0$, what could I say about these sequences? Also, where would the uniform continuity come in? Any help would be appreciated!
Your definition of uniform continuity: $\forall \epsilon > 0$, $\exists \delta > 0$ such that $d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\epsilon$.
$d_X(x_n, x_{n}^{\prime}) \to 0$ means $\forall \epsilon_1 > 0$, $\exists N_1 > 0$ such that $n>N_1 \implies d_X(x_n, x_{n}^{\prime}) < \epsilon_1$.
You need to show $d_Y(f(x_n), f(x_{n}^{\prime})) \to 0$ that is you need to show $\forall \epsilon_2 > 0$, $\exists N_2 > 0$ such that $n>N_2 \implies d_Y(f(x_n), f(x_{n}^{\prime})) < \epsilon_2$.
So let $\epsilon_2 > 0$ be given.
Then for $\epsilon = \epsilon_2$ in the definition of uniform continuity, there is a $\delta_{\epsilon_2} > 0$ such that $d_X(x,y)<\delta_{\epsilon_2} \implies d_Y(f(x),f(y))<\epsilon_2$.
Now for $\epsilon_1 = \delta_{\epsilon_2}$ in the second statement, there is an $N_{\delta} > 0$ such that $n>N_{\delta} \implies d_X(x_n, x_{n}^{\prime}) < \delta_{\epsilon_2}$.
Thus, choosing $N_2 = N_{\delta}$ in the third statement gives you that desired chain of implications $n>N_{\delta} \implies d_X(x_n, x_{n}^{\prime}) < \delta_{\epsilon_2} \implies d_Y(f(x_n), f(x_{n}^{\prime})) < \epsilon_2$.
That is, given any $\epsilon_2 > 0$ you manage to find the corresponding $N_2 > 0$ so that the third statement holds.
Note: Please feel free to edit for sake of readability.