New try:

First observe that since each $\displaystyle f_n$ is continuous at $\displaystyle x$, $\displaystyle f$ is continuous at $\displaystyle x$.

Fix $\displaystyle \epsilon >0$. Now choose $\displaystyle N_1$ such that $\displaystyle n \geq N_1$ implies $\displaystyle | f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}$. (Possible since $\displaystyle f_n \to f$ uniformly.) Next, choose $\displaystyle \delta > 0$ such that $\displaystyle |p-x| < \delta$ implies $\displaystyle | f (p) - f (x) | < \frac{\epsilon}{2}$. (Possible since $\displaystyle f$ is continuous at $\displaystyle x$.) Now, choose $\displaystyle N_2$ such that $\displaystyle n \geq N_2 $ implies $\displaystyle |x_n-x| < \delta$. (Possible since $\displaystyle x_n \to x$.) Let $\displaystyle N$ be the maximum of $\displaystyle \{ N_1 , N_2 \}$. Then $\displaystyle n \geq N$ implies $\displaystyle |x_n-x| < \delta$ which again implies $\displaystyle |f (x_n) - f (x) | < \frac{\epsilon}{2}$, but it also implies $\displaystyle | f_n (x_n) - f(x_n) | \leq \frac{\epsilon}{2}$. But then $\displaystyle | f_n (x_n) - f(x) | \leq | f_n (x_n) - f (x_n) | + | f (x_n) - f(x) | < \epsilon$, so $\displaystyle f_n (x_n) \to f(x)$.

This is me, btw:

$\displaystyle \Rightarrow$