About continuous linear functional on the topology generated by linear functionals

I am self-studying P. Lax's functional analysis book. Here is an exercise in p123. it is supposed to be very easy, but I really couldn't see it.

Could you anyone help me out? Thanks.

Let $\{ l_\alpha\}$ be a collection of linear functions in a linear space $X$ over $\mathbb{R}$ that separates points; that is, for any two distinct points $x$ and $y$ of $X$ there is an $l_\alpha$ such that $l_\alpha(x) \neq l_\alpha(y)$.

Assertion: A linear functional $l$ is continuous in the topology generated by $\{ l_\alpha\}$ iff it is a finite combination of the $l_\alpha$.

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See theorem 3.9 and 3.10 in Rudin's Functional Analysis – Norbert Sep 15 '12 at 13:34
If $l$ is continuous $U = l^{-1}(-1,1)$ is open. Thus there are $l_1,\dots,l_n \in \{l_\alpha\}$ and $\varepsilon \gt 0$ such that $U \supset \bigcap_{i=1}^n \{x\,:\,|l_i(x)| \lt \varepsilon\}$. Deduce that $\ker{l} \supset \bigcap_{i=1}^n \ker{l_i}$ and linear algebra tells you that this means that $l$ is a linear combination of the $l_i$. – t.b. Sep 15 '12 at 13:53
Thank you guys very much! – XxXxX Sep 15 '12 at 13:58
@t.b. : What does $|l_i(x)|\le \epsilon$ mean ?is it not true that $l(x)$ is a discrete point ? Can you help me to get rid of confusion . Thank you . – Theorem Dec 1 '12 at 13:23