Uniform Convergence on Compact Sets Means Uniform Convergence on the whole Set Let $\Omega\in Open(\mathbb{R}^n)$ for some $n\in\mathbb{N}_{\geq1}$.
Then we know that $\exists \{K_n\}_{n\in\mathbb{N}_{\geq1}}$, a collection of compact sets, such that $\Omega = \bigcup_{n\in\mathbb{N}_{\geq1}}K_n$ and $K_n \subseteq int(K_{n+1})\forall n\in\mathbb{N}_{\geq1}$ (in particular, define $K_n := \{x\in\Omega\,|\,||x||\leq n\}\cap\{x\in\Omega\,|\,d(x,\,\Omega^c)\geq\frac{1}{n}\}$).
Suppose there is a sequence of continuous functions $f_i:\Omega\to\mathbb{C}$ which we somehow know converge uniformly to the pointwise limit function $f$ on $K_n$, for all $n\in\mathbb{N}_{\geq1}$.
Prove that $\{f_i\}$ then converges uniformly on $\Omega$.
(This came up while studying Example 1.44 in Rudin's Functional Analysis--it is a fact which he takes for granted and I couldn't figure out why it holds)
 A: I think you are confused by Rudin's claim that the $f_k$ converge to a continuous function. 
What actually happens in Example 1.44 is that Rudin shows that the space $(C^0(\Omega),d)$ is complete, with $d$ the metric he defines in that example. The approach is the same as always: choose a Cauchy sequence, find some element in the space which deserves to be called the limit of that sequence and then show that it actually is. What you seem to have stumbled upon is the second step (find some continuous $f$ which is a candidate for a limit of the Cauchy sequence), in particular the claim that the obvious candidate is continous.
Note that the following is true and easy to see: if $\Omega$ is open in $\mathbb{R}^n$, $K_i\subset \Omega$ a sequence of compact sets such that $K_i\subset K_{i+1}$ for all $i$, $\Omega = \cup_i K_i$ and $f_n:\Omega \rightarrow \mathbb{C}$ is a sequence of continous functions, such that for each fixed $n\in \mathbb{N}$ the sequence $(f_k|K_n)_k$ converges uniformly to some $f:\Omega\rightarrow  \mathbb{C}$, then $f$ is continuous on $\Omega$. The example given in the comments ($x^k:(0,1)\rightarrow \mathbb{R}$) shows that the convergence on $\Omega$ need not be uniform (which you seem to expect).
The proof of the claim is almost trivial: consider some $x \in \Omega$. There will be $n_0$ such that $x\in \mbox{int} K_n$ for $n\ge n_0$, so (by uniform convergence of $(f_k|K_{n_0})_k$) the limit $f$ is continous in $x$. Since $x$ was chosen arbitrarily, $f$ is continous on $\Omega$.
