A proof for the existence of a global minimizer

Question

I wanted to see if my proof of the following theorem is correct.

Theorem. Suppose that $D$, a subset of the Euclidean space, is closed and that all the sublevel sets of the continuous real valued function $f$ on $D$ are bounded. Show that $f$ has a global minimizer.

Proof. Suppose, for the sake of contradiction, that $f$ does not have a global minimizer. So, $\forall x_k\in D$, $\exists \ x_{k+1}$ such that $f(x_{k+1}) \leq f(x_k)$. We can then construct a decreasing sequence $\{f(x_k)\}$ on $D$. Since $f$ and $D$ are bounded, $\exists x^*$ such that $\lim_{k\rightarrow\infty} f(x_k) = f(x^*)$. If $x^*\in D$, we have achieved the desired contradiction. If not, $x^*$ must be on the boundary of $D$. But then since $f$ is decreasing, $\forall x_k$ we can choose $\epsilon <|x_k-x_k^*|$ such that $B_\epsilon(x_k)\subset D$. This contradicts $D$ being closed.

Answer 1

The proof is broken in so many ways

1. $f(x_{k+1})\le f(x_k)$ does not imply anything (after all $(f(x_k))$ can be constant)

2. you get $x_k\to x^*$ and $f(x_k) \to f(x^*)$ only after extracting a subsequence

3. $x^*\in D$ is no contraction to anything

4. $x^*\not\in D$ does not imply $x^*\in \partial D$, $D$ is closed so $\partial D\subset D$

5. $f$ is not decreasing

6. what is $x_k^*$?

7. such an $\epsilon$ does not exist if $x_k$ is itself at the boundary of $D$

8. Wow

You can compare your 'proof' to a proof of Bolzano-Weierstrass. See how such proofs obtain statements that you wanted to get in your attempt.

Edit:

9. why has $(x_k)$ a convergent subsequence? an argument is missing here.