If $f_n$ converges to $f$ uniformly and $f$ is discontinuous at $X$, then $f_n$ is discontinuous at $X$ eventually Hypothesis:

Let $f_n:\mathbb{R} \rightarrow \mathbb{R}$ be a sequence of functions converging uniformly to $f$ on $\mathbb{R}$ (with the same domain). Then if $f$ is discontinuous at $X\in \mathbb{R}$, there exists an $N\in \mathbb{N}$ such that $f_n$ is discontinuous at $X$ for all $n\ge N$

I am almost certain that my hypothesis is false, but I can't see where my proof of the claim goes wrong(or right if you like).
Suppose $f_n$ converges to $f$ uniformly on $\mathbb{R}$. Let $\epsilon > 0$ be arbitrary and $N$ be such that $\forall n\ge N$ it follows $\forall a\in \mathbb{R}$ that
$$f_n(a)\in V_{\epsilon}(f(a))$$
where $V_\epsilon(c)$ is the epsilon neighbourhood around $c$. Suppose $f$ has a discontinuity at $X\in \mathbb{R}$. Then $\forall \delta \gt 0 \exists Y\in V_\delta (X)$ such that
$$ f(Y)\notin V_{3\epsilon}(f(X))$$
A quick sketch of the real line makes these claims apparent(I believe):
$$f_n(X)\in V_\epsilon (f(X))\implies f(Y)\notin V_{2\epsilon}(f_n(X))$$
$$f_n(Y)\in V_\epsilon (f(Y)) \implies f_n(Y)\notin V_{\epsilon}(f_n(X))$$
In conclusion, for any $\delta>0$ around $X$ we can look at $f$ and find a $Y$ where $\epsilon-\delta$ fails, then we can choose $N$ large enough that our $f_N$ and $f$ are wrapped so tightly together that the failure to be continuous holds in $f_N$.
Did I do it?
Edit:
$f_n(x)=\begin{cases} 1/n & \text{if x=}1,1/2,...,1/n \\ x &\text{otherwise}\end{cases}$ might be a counterexample? The hypothesis implies that we can find $N$ such that $f_N$ has at least as many discontinuities as $f$, but in this example, $f_n$ has finitely many discontinuities for all $n$, but $f$ has infinitely many.
 A: You are in the right direction. The key is to notice the a function $f:D\subset\mathbb{R}\rightarrow\mathbb{R}$  is discontinuous at a point $x_0$ iff there exists $\varepsilon_0>0$ such that for $\delta>0$, there are points $x,x'\in(x_0-\delta,x_0+\delta)$ such that
$$|f(x)-f(x')|\geq\varepsilon_0$$
Equivalently,  $f$ is discontinuous at $x_0$ iff  there exists $\varepsilon_0>0$ such that for any $m\in\mathbb{N}$, there are pointa $x_m$ and $x'_m$  such that
$$x_m,x'_m\in\big(x_0-\tfrac1m,x_0+\frac1m\big) \qquad\text{and} \qquad|f(x_m)-f(x'_m)|\geq\varepsilon_0$$
As $f_n$ converges uniformly to $f$, there exists $N\in\mathbb{N}$ such that
$$
|f_n(x)-f(x)|<\tfrac{\varepsilon_0}{4},\qquad\text{for all}\quad x\in D\quad\text{and}\quad n\geq N
$$
For all such $n$,
\begin{align}
\varepsilon_0&\leq |f(x_m)-f(x'_m)|\leq |f(x_m)-f_n(x_m)|+|f_n(x_m)-f_n(x'_m)|+|f_n(x'_m)-f(x'_m)|\\
&\leq \frac{1}{2}\varepsilon_0+|f_n(x_m)-f_n(x'_m)|
\end{align}
Hence
$$
|f_n(x_m)-f_n(x'_m)|\geq\frac{1}{2}\varepsilon_0,\qquad\text{for all}\quad n\geq N
$$
This shows that for all $n\geq N$, the elements $f_n$ of the sequence are discontinuous at $x_0$.
