Let $X$ be a locally compact, $\sigma$-compact Polish (complete and separable metric) space. How to prove: "There is an increasing sequence of continuous cut-off functions with compact support, $0\leqslant f_n(x) \leqslant 1$ and $f_n(x) \to 1$ as $n\to \infty"$?

My proof: let $X=\bigcup_{i=1}^{\infty} K_i$, each $K_i$ is compact. Then we can find a pre-compact neighborhood $U_i \supset K_i$. Using Urysohn's lemma we can find function $g_i=1$ on $K$ and $g_i=0$ on $X \backslash{U_i}$. Let $$ g(x)=g_1(x)+g_2(x)+... $$ If for any point $x\in X$, $x$ belongs to finite number of $U_i$, then this function is well defined. Note that $g(x)\neq 0$ at any point $x\in X$ since $X=\bigcup_{i=1}^{\infty}K_i$. Thus we can define $$ f_1=g_1/g, f_2=(g_1+g_2)/g,..., $$which is what we need.

However, it seems to me that "$x$ belongs to finite number of $U_i$" may not hold. So how to prove this theorem?

And how to relax the assumption on $X$ to get this sequence of functions $f_n$?