I consider the set $S = \{x\in\mathbb{R}^n : \lVert x\rVert < 1\}$ and $f : S\to\mathbb{R}$ a continuous function satisfying
$$f(x)\geq\frac{1}{1-\lVert x\rVert}\quad (1)$$
The problem I want to solve is
$$\inf_{x\in S}f(x)\quad (Q)$$
We can prove that $Q$ has a finite value using $(1)$ since it is bounded below. Now I would like to use a method of "exhaustion" that consists to consider the following sequence of problem for $k\geq 1$ :
$$\inf_{x\in S_k}f(x)\:\text{with}\: S_k =\{x\in\mathbb{R}^n : \lVert x\rVert \leq 1 - \frac{1}{k}\}\quad (Q_k)$$
and observe that by continuity of $f$ and compactness of the $S_k$ we have a solution to the problem $Q_k$. The next step is to consider a sequence $x_k$ defined as follows :
$x_1$ is a solution of $Q_1$
for $k\geq 2$, $x_k = x_{k-1}$ if $x_{k-1}$ is a solution of $Q_k$, otherwise $x_k$ is any solution of $Q_k$
It can be shown that it implies that the sequence $\lVert x_k\rVert$ is increasing. Now by compactness of the unit ball, we can extract a convergent subsequence, lets say $\bar{x}_k$ with limit $\bar{x}\in B(0,1)$. We observe two things :
the subsequence still have the property that the sequence of norms is increasing
the sequence $f(\bar{x}_{k})_{k}$ is decreasing.
This implies that $\lVert\bar{x}\rVert < 1$ : otherwise, using $(1)$ we get that the sequence $(f(\bar{x}_k))_k$ goes to $\infty$, so letting $M>f(\bar{x}_1)$ we have that for some $N\in\mathbb{N}$ :
$$f(\bar{x}_k)\geq M>f(\bar{x}_1), \forall k\geq N$$
which contradicts the second observation. Hence $\bar{x}\in S$ and by continuity $(f(\bar{x}_k)_k$ converges, decreasingly, to $f(\bar{x})$. Now let $y\in S$, we have that $\lVert y\rVert < 1$, this implies the existence of some $k_0\in\mathbb{N}$ such that $y\in S_{k_0}$, hence we get
$$f(y)\geq f(\bar{x}_{k_0})\geq f(\bar{x})$$
This proves that problem $Q$ has a solution.
I would like to know if it is correct please. I think the main ideas are here but I would like to have some advice to know if it is clear enough please.
Thank you a lot !