First, a note on notation: $\inf_{B_{\delta'}(x)} \delta'$ doesn't really make sense (although it is clear what you meant to say). You should make explicit the dependence of $\delta'$ on $x$, e.g. by saying "As this is an open cover of $S$, around each $x \in S$ there exists some real number $\delta'(x) > 0$ such that the open ball $B_{\delta'(x)}(x)$ is contained in some element of the subcover". Then you can define $\delta = \inf_{x \in S} \delta'(x)$.
In any case, the answer to your question is (as stated) no. The issue is that the numbers $\delta'(x)$ are not required to "try to be large"! Indeed, for each $x \in S$, there are arbitrarily small positive real numbers $r$ such that $B_r(x)$ is contained in some element of the subcover. This allows us to pick values for $\delta'$ such that $\delta = 0$, as in the following example.
Example. Let $X = \mathbb{R}$ and $S = [0,1]$. Consider the open cover $\{(-1,2)\}$. Choose any bijection $f : S \to (0,1)$. For each $x \in S$, let $\delta'(x) = f(x)-f(x)^2$. Note that $0 < \delta'(x) \leq \frac{1}{4}$ for all $x \in S$, and consequently $B_{\delta'(x)}(x) \subseteq (-1,2)$ for all $x \in S$. However,
$$\delta = \inf_{x \in S} \delta'(x) = \inf_{y \in (0,1)} (y - y^2) = 0.$$
Hopefully this example makes it clear that it's easy to cook up examples where $\delta'$ is chosen poorly, resulting in $\delta = 0$. However, there is a more interesting fact: it is always possible to pick values for $\delta'$ such that $\delta > 0$.
Theorem. Let $(X,d)$ be a metric space and let $S \subseteq X$ be compact. Let $U_1, \dots, U_n$ be open subsets of $X$ such that $\bigcup_{i=1}^n U_i = S$. Then there exists a function $\delta : S \to (0,\infty)$ such that:
- For all $x \in S$, there is some $1 \leq i \leq n$ such that $B_{\delta(x)}(x) \subseteq U_i$.
- $\inf_{x \in S} \delta(x) > 0$.
Note: This is a simplified form of the Lebesgue Number Lemma, which answers your strengthened question. The Lebesgue Number Lemma is pretty easy to prove, though, so what follows is a modified proof that applies in this context.
Proof. First, suppose $U_i = X$ for some $i$. Then we may choose $\delta(x) = 1$ for all $x \in S$ and we are done. Otherwise, we have that $Z_i := X \setminus U_i$ is nonempty for all $1 \leq i \leq n$. Let $f : S \to (0,\infty)$ be defined by $f(x) = \frac{1}{n} \sum_{i=1}^n d(x,Z_i)$, where $d(x,Z_i)$ means $\min_{z \in Z_i} d(x,z)$ (note: this is always non-negative). To prove that $f$ is well-defined, let $x \in S$ be arbitrary. There is some $1 \leq i \leq n$ such that $x \in U_i$, so $x \notin Z_i$, whence $d(x,Z_i) > 0$, so $f(x) > 0$. Now, to verify property 2, we simply note that $f$ is continuous (it is the sum of continuous functions), hence (by compactness of $S$) attains a minimum value $D > 0$. Finally, let $\delta : S \to (0,1)$ be defined by $\delta(x) = D$ for all $x \in S$. Since $\delta$ is constant and positive, property 2 is trivially satisfied. To check property 1, let $x \in S$ be arbitrary. Since $f(x) \geq D$, there is some $1 \leq i \leq n$ such that $d(x,Z_i) \geq D$. This means that $D_{\delta(x)}(x) = B_D(x) \subseteq U_i$.
Best Answer
As stated in my comment, $\forall (x,y) \in U$, take $r = \frac{|y|}{2}$. Show $B_r((x,y)) \subset U$.
For $(x', y') \in B_r((x,y))$,
Then $|y'| > |y| - \frac{|y|}{2} > 0 \implies (x',y') \in U$