Inequality involving functions bounded above but not below: $|\sup_{x\in\mathbb R}f(x)-\sup_{x\in\mathbb R}g(x)|\leq\sup_{x\in\mathbb R}|f(x)-g(x)|$?

Let $f,g: \mathbb{R} \rightarrow \mathbb{R}$ be bounded above and below, then we can prove that: $$| \sup_{x \in \mathbb{R} } f(x) - \sup_{x \in \mathbb{R} } g(x) | \leq \sup_{x\in \mathbb{R}} | f(x) - g(x) |$$ Suppose that $f,g$ are bounded above, but not below, and that $| f - g |$ is bounded. Then, by the completeness of $\mathbb{R}$, we know that $$\sup_{x\in \mathbb{R}} | f(x) - g(x) | \in \mathbb{R}$$ I would like to prove that the inequality still holds. If this is true then we can generalize and say that the inequality holds for all $f,g$ bounded above ( since in the cases when only one of the two functions is bounded or $|f - g|$ is not bounded, $\sup_{x\in \mathbb{R}} | f(x) - g(x) |$ is neither bounded).

Thank you for your time, any advice will be of great help.

Let $(x_n)_n$ be a sequence such that $\lim_n f(x_n)=\sup\limits_{x\in\mathbb R}f(x)$. Then $\limsup_n \left(f(x_n)-g(x_n)\right)=\sup\limits_{x\in\mathbb R}f(x)-\liminf_n g(x_n)$. For each $n$, $g(x_n)\leq \sup\limits_{x\in \mathbb R}g(x)$, so $\liminf_n g(x_n)\leq \sup\limits_{x\in \mathbb R}g(x)$. Thus we have
\begin{align*} \sup_{x\in\mathbb R} f(x)-\sup_{x\in\mathbb R} g(x)&\leq \sup_{x\in\mathbb R} f(x) - \liminf_n g(x_n)\\ &=\limsup_n\left(f(x_n)-g(x_n)\right)\\ &\leq \sup_{x\in\mathbb R} \left(f(x)-g(x)\right)\\ &\leq \sup_{x\in\mathbb R}|f(x)-g(x)|. \end{align*}
By swapping the roles of $f$ and $g$ we get $$\sup_{x\in\mathbb R} g(x)-\sup_{x\in\mathbb R} f(x)\leq \sup_{x\in\mathbb R}|f(x)-g(x)|,$$ and therefore
$$\left|\sup_{x\in\mathbb R} f(x)-\sup_{x\in\mathbb R} g(x)\right|\leq \sup_{x\in\mathbb R}|f(x)-g(x)|.$$