Suppose that $f(x)$ is a decreasing function in $x$, and $f(x) \leq c$ for a constant $c$. I am trying to figure out if it is true that $x \geq k$ for another constant $k$?
Trivially, if we are told that $f$ is invertible, then I have that:
$$ f(x) \leq c \implies x \geq f^{-1}(c) = k $$
However, does the relation hold even if $f$ is not invertible?
Since $f$ is decreasing you have $x\ge k\Rightarrow f(x)\le f(k)$.
So your relation is true for any $k$ such that $f(k)\le c$.
The case when $f$ is continuous and strictly decreasing and $\lim\limits_{x\to -\infty} f(x)>c$ ensures there always exists an unique $k$ such that $f(k)=c$.
But as stated by Wnitram and Florence if the limit in $-\infty$ is less than $c$, then there exists no $k$ such that $f(k)=c$ exactly but it is true that $f(k) For the case where the limit in $-\infty$ is $>c$ but $f$ is discontinuous (for instance $f(x)=-\lfloor x\rfloor$), it is possible that there is no $k$ such that $f(k)=c$, but still if there exists an $x_0$ such that $f(x_0)\le c$ then $k=x_0$ is suitable, yet not necessarily the "best" one. I think that in your question you are mistaken : For instance if you consider $f(x)=-\lceil x\rceil$, and $c=-1$. $f(]-1,0])=\{0\}$ $f(]0,1])=\{-1\}$ $f(]1,2])=\{-2\}$ You have $k_{inf}=0$ but $f(0)=0>c$ therefore $k_{min}$ does not exists, yet $k=0.01$ is a suitable constant, but not the "best". If $f$ is strictly decreasing the invertibility of $f$ at $c$ ensures that $k_{min}=f^{-1}(c)$ exists. If $f$ is constant (so decreasing but not strictly), $k_{min}$ may not exists. For instance consider $\forall x,f(x)=c$. In general $k_{min}$ may or may not exists, but that does not prevent one to find a suitable $k$ even if not the best, since it comes from your hypothesis that $f(x_0)\le c$ for some $x_0$. Hope it is clearer.
No, this is not always the case. $f(x) = -\tan^{-1}x$ is a counterexample. We have $f(x) \leq \frac{\pi}{2} = c$; however, your proof fails, because $f^{-1}(c) = \tan(c)$ is not defined.