Error in uniform convergence question? 
Suppose that $(X,\ d)$ and $(Y,\ \rho)$ are metric spaces, that
  $f_n:X\to Y$ is continuous for each $n$, and that $(f_n)$ convergence
  pointwise to $f$ on $X$. If there exists a sequence $(x_n)$ in $X$
  such that $x_n\to x$ in $X$ but $f_n(x_n)\not\to f(x)$, show that
  $(f_n)$ does not converge uniformly to $f$ on $X$.

I've managed to "prove" that the question is self-contradictory, so please find my error.
$\forall n\in\mathbb{N}$, $f_n$ is continuous at $x$. Let $\epsilon>0$. $\forall n\in\mathbb{N}$ $\exists\delta>0$ s.t. $\rho(f_n(y),\ f_n(x))<\epsilon/2$ for all $y\in X$ s.t. $d(y,\ x)<\delta$. ------------ (1)
Since $x_m\to x$, $\exists N_1\in\mathbb{N}$ s.t. $d(x_m,\ x)<\delta$ $\forall m\ge N_1$. ------------- (2)
From (1) and (2),
$\forall n$, $\forall m\ge N_1\implies d(x_m,\ x)<\delta\implies \rho(f_n(x_m),\ f_n(x))<\epsilon/2$ ------------- (3)
Also, $f_n(y)\to f(y)$ for all $y\in X$ due to pointwise convergence. So, $\exists N_2\in\mathbb{N}$ s.t. $\rho(f_n(y),\ f(y))<\epsilon/2$ $\forall n\ge N_2$ and $\forall y\in X$. ----------- (4)
Let $N_3=\max(N_1,\ N_2)$. Suppose $n\ge N_3$. Then $n\ge N_1$ and $n\ge N_2$.
$\begin{aligned}
\implies\rho(f_n(x_n), f(x))&\le\rho(f_n(x_n),\ f_n(x))+\rho(f_n(x),\ f(x)) \\
&<\epsilon/2+\epsilon/2\text{ [Using (3) and (4)]} \\
&=\epsilon
\end{aligned}$
I have thus "proved" that $f_n(x_n)\to f(x)$, contradicting the question. Where have I gone wrong?
 A: If we are more precise, we may see where the error is. I will use $|y-x|$ to mean $d(y,x)$ since I think it makes it clearer than working with $d$ and $\rho$. It won't change any of the important details. 
For each $n$, $f_n$ is continuous at $x$. Let $\epsilon>0$. Then, for each $n\in\Bbb N$, there is a $\delta = \delta(n)$, which depends on $n$, such that $|f_n(y)-f_n(x)|<\epsilon/2$ whenever $|y-x|<\delta(n)$.
Since $x_m\to x$, there is an $N_1 = N_1(n)\in\Bbb N$ (note that $N_1$ also depends on $n$ here!) such that $|x_m - x| < \delta(n)$ for each $m\ge N_1$.
Now you claim that for each $k,m\ge N_1(n)$, whenever $|x_m-x|<\delta(n)$, you have $|f_k(x_m)-f_k(x)| < \epsilon/2$.
This is the first serious point you erred. The reason this is an incorrect deduction is that for different $n$, you don't know that $N_1 = N_1(n)$ is large enough to guarantee that $|f_k(x_m)-f_k(x)|<\epsilon/2$. Only for the specific $n$ for which $\delta = \delta(n)$ is this true.

Edit: RRL summed it up nicely in the comment. You are assuming equicontinuity of the sequence $\{f_n\}$ when you are neglecting the $n$ that the $\delta = \delta(n)$ depends on.
A: This is an excellent example of how imprecise language/notation can lead to confusion. 
In your first line you write "for all $n$, there exists $\delta$ such that..." This sort of suggests that the same $\delta$ can be used for all $n$, which is not correct and is what leads to your conclusion.
In the future it might be more useful to think of this as "for each $n$, there exists $\delta$...". At least to me, this suggests more strongly that the $\delta$ may be different for each $n$.
A: For a correct proof (by contradiction), show that $f(x_n) \not\to f(x)$ is impossible if convergence is uniform using
$$\rho(f_n(x_n), f(x)) \leqslant \rho(f_n(x_n), f(x_n)) + \rho(f(x_n), f(x))$$
Note that $f$ must be continuous when the sequence of continuous functions $f_n$ is uniformly convergent. Thus each term on the RHS is smaller than $\epsilon/2$ for sufficiently large $n$  -- the first by the uniform convergence $f_n \to f$ and the second by the convergence $x_n \to x$ and the continuity of $f$. 
