We assume that we know the Least Upper Bound Principle: Every non-empty set $A$ which is bounded above has a least upper bound.
We want to show that every non-empty set $B$ which is bounded below has a greatest lower bound.
Let $A=-B$. So $A$ is the set of all numbers $-b$, where $b$ ranges over $B$.
First we show that if $B$ has a lower bound, then $A$ has an upper bound. For let $w$ be a lower bound for $B$. We show that $-w$ is an upper bound for $A$.
Because $w$ is a lower bound for $B$, we have $w\le b$ for any $b\in B$. Multiply through by $-1$. This reverses the inequality, and we conclude that $-w\ge -b$ for any $b\in B$. The numbers $-b$ are precisely the elements of $A$, so $-w\ge a$ for any $a\in A$.
Since $A$ is bounded above, it has a least upper bound $m$. We show that $-m$ is a greatest lower bound of $B$.
As usual, the proof consists of two parts (i) $-m$ is a lower bound for $B$ and (ii) nothing bigger than $-m$ is a lower bound for $B$.
The proofs again use the fact that multiplying by $-1$ reverses inequalities. To prove (i), suppose that $-m$ is not a lower bound for $B$. Then there is a $b\in B$ such that $b\lt -m$. But then $-b\gt m$. Since $-b\in A$, this contradicts the fact that $m$ is an upper bound of $A$.
To prove (ii), one uses the same strategy.
You say that proving the last claim is easy, but in fact that is where your problem is. (That's a good rule of thumb, by the way - if you are tempted to just say the proof is obvious, write it out anyway because it might not be!)
The trouble is that all you know is that $I = \displaystyle \bigcap_n I_n$ is nonempty - it may have more than one element, and for all you know, it might even contain elements of $S$! For instance, take $S = \{1-\frac{1}{n}\ |\ n\in \mathbb{N}\} \cup \{2-\frac{1}{n}\ |\ n\in \mathbb{N}\}$ and note that your procedure might end up with $I = [1, 3]$.
Best Answer
First of all note that the ideas in your question constitute almost 99% of the proof and handle the most difficult parts of the proof. And the remaining part of the proof is obtained by getting contradictions for assumptions $f(c) > v$ and $f(c) < v$ forcing us to conclude that $f(c) = v$. The contradiction follows from the following local property of continuous functions (it also goes by the name of sign preserving property):
If $f$ is continuous at $a$ and $f(a) \neq 0$ there is a neighborhood of $a$ in which $f$ maintains the same sign as that of $f(a)$.
This is an important but easy consequence of definition of continuity and I hope you can prove this by yourself.
Now consider your question and assume that $f(c) > v$. Then using the above sign preserving property we can prove that there is a neighborhood $I$ of $c$ such that all values of $f$ in $I$ are greater than $v$. Since $x_{n} \leq c \leq y_{n}$ and $$\lim_{n \to \infty}x_{n} = \lim_{n \to \infty}y_{n} = c$$ it follows that there is a value of $n$ for which $[x_{n}, y_{n}] \subseteq I$ and since $f(y_{n}) \leq v$ gives us a contradiction (as $y_{n} \in I$ and hence $f(y_{n}) > v$).
Update: The question has changed and the given answer corresponds to original version where it is asked to establish intermediate value theorem via the use of the Nested interval principle.