Question: I am having a hard time finding such an example. This may be a simple question but I am very new to analysis. Finding such $\{a\_n\}$, where $\{a\_{2k}\}$ converges to a number seems easy and one such example could be $\{(-1)^n\}$. In addition, if they asked for such sequences where two subsequences converge to different numbers sine may help us, but this question is a bit out of my range or probably I am missing something super basic.
Can I get some help, please!

Answer: 
Following from spaceman's comment, let
$$F(s)=\int\_0^1f(x+s)dx.$$
Then $F'(s)=f(s+1)-f(s)>0$, since $f$ is strictly increasing. Further, we also have the bounds
$$f(s+1)\geq F(s)\geq f(s)$$
We conclude that $F(x)$ is also a strictly increasing continuous bijection $\mathbb{R}\rightarrow I$. Now let $\alpha\in I$. Then for each $b\in\mathbb{R}$ with $F(b)<1-\alpha$, there is a unique $a\in\mathbb{R}$ such that $F(a)-F(b)=\alpha$ given by $a=F^{-1}(F(b)+\alpha)$. There are uncountably many choices for $b\in \mathbb{R}-\mathbb{Q}$, and only countably many choices for $a\in\mathbb{Q}$, so there must be some choice of $a$ and $b$ both irrational.
@winic92046 This is too long for a comment, so have written it here. But know that this is just a comment for now.

A train of thought:

First, given any real number $\alpha\in(0,1)$, it is [possible](https://math.stackexchange.com/questions/2743302/prove-that-any-real-number-can-be-expressed-as-the-sum-of-two-irrational-numbers) to show that there exist two irrationals $\beta, \gamma$ such that $\alpha = \beta + \gamma$.

@winic92046 A train of thought:

First, given any real number $\alpha\in(0,1)$, [there exist]]([Prove that any real number can be expressed as the sum of two irrational numbers](https://math.stackexchange.com/questions/2743302/prove-that-any-real-number-can-be-expressed-as-the-sum-of-two-irrational-numbers)) two irrationals $\beta, \gamma$ such that $\alpha = \beta + \gamma$.

Now, if we can show that $F\vert\_{\mathbb{R}\backslash\mathbb{Q}} : \mathbb{R}\backslash\mathbb{Q} \to \mathbb{R} $ is surjective on some semi-infinite space of irrationals (or reals), we could continue by working backwards.

In essence, we could continue by showing that for any irrational $\beta$, there exists an irrational $a$ such that $\beta = F(a)$. Similarly, for $\gamma$, there exists an irrational $b$ such that $\gamma = F(b)$. From here, we would then be able to conclude the claim.

I don't imagine it will be quite as simple as this, as there are some details that I neglected, but I believe a similar approach is worth trying.