Prove that the CDF of a random variable is always right-continuous

Let $X$ be a random variable with cumulative distribution function $F_X$. It is a known fact that this function $F_X$ is right-continuous. But I'm having some trouble to prove this result. Below I'm sketching out what I have done so far to prove this and where exactly I'm having the difficulty.

My attempt:

To show that $F_X$ is right-continuous at a real number $y$, we need to prove that $$\lim_{x\to y^+}F_X(x)=F_X(y).$$

I will use the sequential approach to prove this. Let $(x_n)$ be a sequence such that $x_n \to y$ and $x_n \geq y$ for every $n$.

Define $A_n = (-\infty,x_n]$ for each $n$. Now from the sequence $(x_n)$ we can extract a monotonically decreasing subsequence, say $(x_{n_k})$. As the new subsequence is monotonically decreasing, it follows from our definition of $A_n$'s that for all $k$ in $\mathbb{N}$, $A_{n_{k+1}} \subseteq A_{n_k}$. So $X^{-1}(A_{n_{k+1}}) \subseteq X^{-1}(A_{n_k})$ and hence $$F_X(x_{n_{k+1}}) = P(X\leq x_{n_{k+1}}) = P(X^{-1}(A_{n_{k+1}})) \leq P(X^{-1}(A_{n_{k}})) = P(X\leq x_{n_{k}}) = F_X(x_{n_{k}}).$$

Now since $(X^{-1}(A_{n_k}))_k$ is a nested sequence of events, we know $$\lim_{k\to\infty}P(X^{-1}(A_{n_k})) = P(\cap_{k=1}^{\infty}(X^{-1}(A_{n_k}))) = P(X^{-1}(\cap_{k=1}^{\infty}A_{n_k})) = P(X^{-1}(-\infty,y]).$$

Therefore, $$\lim_{k\to\infty}F_X(x_{n_k}) = P(X\leq y) = F_X(y).$$

This shows that the sequence $(F_X(x_{n_k}))_k$ converges to $F_X(y)$. But it does not prove $F_X(x_n) \to F_X(y)$ as $n\to \infty$, which is required to show that $F_X$ is right-continuous. I'm completely stuck at this point and can't get any idea how to go from the particular subsequence $(F_X(x_{n_k}))_k$ to the general sequence $(F_X(x_n))_n$ or how to prove this result using the sequential approach and without having to extract a monotone decreasing subsequence from a given sequence.

I'll appreciate any help or suggestion. Thanks and regards.

Your struggle comes from a lack of understanding of continuity. In order to show $f:\mathbb{R}\to\mathbb{R}$ is (left/right)-continuous, it suffices to show that if $x_n \to x$ monitonically (from the left/right) that $f(x_n) \to f(x)$.
But anyway, your argument can be completed as follows. Choose $A_n=\left(-\infty,x+\frac{1}{n}\right]$ as your intervals. As you said, you can prove $F_X\left(x+\frac{1}{n}\right) \to F_X(x)$. Now , for each $\epsilon$, you can find $n$ such that: $$0 \leq F_X\left(x+\frac{1}{n}\right)-F_X(x)<\epsilon.$$ Because the CDF is monotonic, choose $\delta<\frac{1}{n}$. Then for all $y$, with $x<y<x+\delta$ we have: $$0 \leq F_X(y)-F_X(x)\leq F_X(x+\delta)-F_X(x)\leq F_X\left(x+\frac{1}{n}\right)-F_X(x)<\epsilon,$$ which finishes the proof.