Continuous functions on intervals. Problem Statement:$$
\text{Let}\ I = [a,b]\ \text{and let}\ f: I \rightarrow \mathbf{R}\ \text{and}\ g: I \rightarrow \mathbf{R}\ \text{be continuous on}\ I.\\ \text{Show that the set}\ E := \{x \in I: f(x) =g(x)\}\ \text{has the property that if}\ (x_n) \subseteq E\ \text{and}\ x_n \rightarrow x_0,\ \text{then}\ x_0 \in E. $$
Here's what I've got so far: (I'll mark the part in red that I want more clarification on)$$
\text{Since}\ (x_n) \in E \Rightarrow f(x_n) = g(x_n).\ \\ \text{Since}\ f,g\ \text{are continuous on}\ I \Rightarrow lim_{n \rightarrow 0} f(x_n) = f(x_0)\ \text{and}\ lim_{n \rightarrow 0} g(x_n) = g(x_0).\ \\ {\color{Red} {\text{Since}\ f(x)=g(x)\ \forall x \in E\ \text{and}\ (x_n) \in E \Rightarrow f(x_0) = g(x_0).}}     
$$
Any feedback is welcome. I think I'm pretty close to a correct proof, I may just need one extra piece of information that I'm missing. If this is a correct and sufficient proof, can someone please explain the logic in the last line for me. Thanks in advance. 
 A: You probably mean $\lim \limits_{n\to \color{red}{+\infty}}$. 
Do not use $\implies$ to abbreviate 'then'. Choose between $P\implies Q$ or 'if $P$, then $Q$', do not mix it with 'If $P\implies Q$' - this means something else. 
I don't really understand your red statement. 
Here's how I'd write your proof: 
Since for all $n\in \Bbb N$, it holds that $x_n\in E$, then $f(x_n)=g(x_n)$. (What comes next is simply a rigorous treatment of taking the limit in the last equality). Now define two new sequences $(y_n)_{n\in \Bbb N}$ and $(z_n)_{n\in \Bbb N}$ by $y_n=f(x_n)$ and $z_n=g(x_n)$, for all $n\in \Bbb N$.
Clearly the following limits exist and the equalities hold: $$\lim \limits_{n\to +\infty}(y_n)=\lim \limits_{n\to +\infty}(f(x_n))\color{green}=f\left(\lim \limits_{n\to +\infty}(x_n)\right)=f(x_0)$$
and similarly $$\lim \limits_{n\to +\infty}(z_n)=\lim \limits_{n\to +\infty}(g(x_n))\color{green}=g\left(\lim \limits_{n\to +\infty}(x_n)\right)=g(x_0).$$
The green equalities hold due to continuity of $f$ and $g$, but you should think about why is it that $x_0\in I$, (so that $f(x_0)$ and $g(x_0)$ make sense).
Now, since $y_n=z_n$ for all $n\in \Bbb N$, then the sequences $(y_n)_{n\in \Bbb N}$ and $(z_n)_{n\in \Bbb N}$ are the same, so they have the same limit, that is $f(x_0)=g(x_0)$, hence $x_0\in E$.
A: Hint: $E$ is the set of the zeros of the continuous function $h=f-g$, that is, $E$ is the preimage of a closed set (what set?) by a continuous function. So $E$ is...
A: You are almost correct instead of small typos (like we should have $n \to \infty$ instead of $n \to 0$) We have $\lim_{n \to \infty}f(x_{n}) = \lim_{n \to \infty}g(x_{n})$ and since $f, g$ are continuous the limit operation can be exchanged with functional operation and we get $f(\lim_{n \to \infty}x_{n}) = g(\lim_{n \to \infty}x_{n})$ i.e. $f(x_{0}) = g(x_{0})$ and hence $x_{0} \in E$.
