Countable squence bounding all continuous funcitons uniformly? This is rephrasing of this question.  Does there exist a sequence of point-wise continuous functions $\{f_n\}_{n=1}^{\infty}$ in $C(\mathbb{R}^k,[0,\infty))$ satisfying
$$
f_n(x)\leq f_n(z) \mbox{ if } \|z\|\geq \|x\|
$$
such 
that for any $g \in C(\mathbb{R}^k,\mathbb{R}^l)$ there exists some $N>0$ such that
$$
\sup_{x \in \mathbb{R}^k} \|f_N(x)g(x)\|<\infty?
$$
 A: If zero values of $f_n$ are allowed then there is an obvious positive answer. Pick some $R>0$ and let $f_1$ be an arbitrary function satisfying the condition such that $f_1(x)=0$  provided $\|x\|\ge R$. For instance, we can put $f_1(x)=\max\{R-\|x\|,0\}$.
If zero values of $f_n$ are not allowed then the answer is negative even for $k=l=1$. Indeed, given any sequence $\{f_n\}\in C(\Bbb R,(0,\infty))$ for each $n\in\Bbb Z$ put $g(n)=|n|/\min_{i\le n} f_i(n)$ and then extend the function $g$ to $\Bbb R$ piecewise-linearly on each interval $[n,n+1]$. Then for each $N$ we have $$\lim_{n\to +\infty} f_N(n)g(n)\ge \lim_{n\to +\infty} f_N(n)n/f_N(n)=+\infty.$$ 
I expect it is easy to show that the minimum cardinality of the family $f_n$ bounding all continuous function equals to a small cardinal $\frak d$, well-known in the Modern Set Theory, see [JW], [Va]. It is defined as the smallest size of a set $D$ of functions from $\omega$ to $\omega$ such that for each $g:\omega\to\omega$ there exists a function $f\in D$ such that $f(n)\ge g(n)$ for each $n\in\omega$. Is it easy to see that $\omega_1\le\mathfrak d\le\mathfrak c$. The Martin Axiom implies $\mathfrak d=\mathfrak c$. On the other hand, there are models of ZFC with $\mathfrak d<\mathfrak c$, see [Va]. 
References
[JW] W. Just, M. Weese. Discovering Modern Set Theory. II, Graduate Studies in Math. 18, Providence: AMS, 1997.
[Va] J. E. Vaughan. Small uncountable cardinals and topology // in: Open Problems in Topology, ed: J. van Mill and G.M.Reed, Amsterdam: North-Holland, 1990, 195–216.
