The construction of function If $X$ is a paracompact Hausdorff space and $\{ {X_i}\} $ $i \ge 0$ are open subsets of $X$ such that ${X_i} \subset {X_{i + 1}}$ and $\bigcup\nolimits_{i \ge o} {{X_i}}  = X$, can we find a continuous function $f$ such that $f(x) \ge i + 1$ when $x \notin {X_i}$ ?

Yes, your function exists.
As $X$ is paracompact and Hausdorff you get a partition of unity subordinate to your cover $\{X_i\}$, that is, a family $\{f_i : X \rightarrow [0,1]\}_{i \geq 0}$ with $\text{supp}(f_i) \subset X_i$ and every $x \in X$ has a neighborhood such that $f_i(x) = 0$ for all but finitely many $i \geq 0$ and $\sum_i f_i(x) = 1$.
That implies that $g = \sum_i i \cdot f_i$ is a well defined, continuous map. For $x \notin X_k$ you have $f_i(x) = 0$ for all $i \leq k$ and that implies
$$ \sum_{i \geq 0} i \cdot f_i(x) = \sum_{i \geq k+1} i \cdot f_i(x) \geq (k+1) \sum_{i \geq k+1} f_i(x) = k+1, $$
that is, $g(x) \geq k+1$ for all $x \notin X_k$.
