How to prove that $X \cap Y$ is closed? I have to prove that $X \cap Y$ is closed. 
My idea is to apply the definition of a closed set: 
Consider a sequence $(x_n)$ of members of $X \cap Y$. If the sequence $x_n$ tends to some limit point $x$ in $\mathbb{R}$ as $n$ tends to infinity, then we need to show that $x$ is a member of the set $X \cap Y$. That will prove that $X \cap Y$ is closed. 
This idea is correct? or, can we prove it with another method?
 A: As has been pointed out in some comments, you have to say with which topology you want to work. Unfortunately your response to a comment is illegible to me. 
The space of continuous functions is typically equipped with the topology arising from the sup norm,
$$ ||f|| = \sup\{|f(x)|\, x \in [a,b]\} $$
(assuming you are looking at continuous functions on $[a,b]$). Since the space of continuos functions, with this norm, is a Banach space, the question whether a set is closed can be answered by looking at sequences, as you suggested. 
In general, if your topology is derived from a metric, the same conclusion is valid, see, e.g., here:
Closed set in normed vector space
A: As Thomas has pointed out, to prove that a set is closed in a metrisable topology it is fine to prove that it is sequentially closed. Thus it sufficies to show that for every sequence $(f_n)\subset Y$ that converges to $f\in X$ we must have $f\in Y$.
For any such sequence we have
$$\|f-f_n\|_X=\sup_{x\neq y}\frac{|f_n(x)-f_n(y)-(f(x)-f(y))|}{|x-y|}\to 0.$$
In particular
$$\frac{|f_n(1)-f_n(0)-(f(1)-f(0))|}{1}=|f_n(1)-f(1)|\to 0$$
which means $f(1)\leq1$. Thus $f\in Y$. As $(f_n)$ and $f$ are arbitrary $Y$ is closed.
