If $f:B\to\mathbb{R},\ B\subset\mathbb{R}$ is increasing then there is a sequence of strictly increasing functions whose pointwise limit is $f$ I have proved the following statement and I would like to know if my proof is correct and/or how it could be improved.

Suppose $B\subset\mathbb{R}$ and $f:B\to\mathbb{R}$ is an increasing function. Prove that there exists a sequence $f_1, f_2,\dots$ of strictly increasing functions from $B$ to $\mathbb{R}$ such that $f(x)=\lim_{k\to\infty}f_k(x)$ for every $x\in B$.

My proof:
The sequence defined by $f_k(x):=f(x)+\frac{x}{k},\ k\geq 1, x\in B$ satisfies the requirements for if we let $k\geq 1$ and $x,y\in B,\ x<y$ then $f_k(y)-f_k(x)=f(y)-f(x)+\frac{y-x}{k}>0$ so $f_k(y)>f_k(x)$ ie $f_k$ is strictly increasing and also, for any $x\in B$ we have $\lim_{k\to\infty}f_k(x)=\lim_{k\to\infty}(f(x)+\frac{x}{k})=\lim_{k\to\infty}f(x)+\lim_{k\to\infty}\frac{x}{k}=f(x)+0=f(x)$. $\square$
 A: Looks fine to me. We can generalise this in an obvious way. Let $g : B \to \Bbb R$ be any strictly increasing function. Then, $$f_k := f + \frac{g}{k}$$
does the job.
Note that in your example, we don't have any comparison between $f_k$ and $f$, since your $g$ is not of one sign. So we have $f_1(x) < f_2(x) < \cdots < f(x)$ for some $x$ and the reverse for other.
However, what one could do is take a $g$ which is say, strictly negative. (For example, $g(x) = -e^{-x}$.) In this case, we have $$f_1 < f_2 < \cdots < f$$ and $f_k \to f$ pointwise.
This could be more convenient for some applications. (The reason I say this is because this reminds me of how one approximates measurable functions by simple functions from below in measure theory.)

Edit: This is based on Yorch's nice comment. They've suggested trying to make the convergence uniform.
Again, putting one more condition on $g$ can ensure that. Can you see what it is? (As a hint, it may be useful to recall that $f_k \to f$ uniformly iff $\sup_B |f_k - f| \to 0$. The sequence here would be $\frac1k\sup_B |g|$.)
