Ok, I am going to mix and match formality and informality, here. Because I can. We are told from the outset that $f$ is continuous. In particular, this means it is continuous at $c$. What is one formal way of saying $f$ is continuous at $c$? We say that $\lim\limits_{x \to c} f(x) = f(c)$.
I am going to call $f(c)$, $u$. Why? I like that letter (Don't you?). We are also told that $f(c) = u > 0$, in short: it is positive. Now, when we say that $f$ is continuous at $c$ (that limit business), what we mean is that if $x$ is near $c$, $f(x)$ is near $u$. So surely if we pick some $x$ close enough to $c$, we ought to have $f(x)$ close enough to $u$ that it is still positive (even if it's just a little bit less than $u$, values larger than $u$ will be, of course, even more positive).
How close is "close enough"? Here the formal definition of a limit comes to our rescue. When we say $\lim\limits_{x \to c} f(x) = f(c) = u$, what we mean (formally) is given any positive number (even an itty bitty one), which we call $\epsilon$ (apparently, Greek letters are way cooler than English ones), we can find some positive number $\delta$ so that whenever |x - c| < \delta, then |f(x) - u| < \epsilon. Since we already know $f$ is continuous, we get "$\delta$" for free, anytime we pick an "$\epsilon$". So in this case, "close enough" means "within $\delta$".
Now, all we need to do, is choose $\epsilon$. Following the steps of the great set-theorist Indiana Jones, we shall "choose wisely". Our choice? We shall use $\epsilon = \frac u 2$. Our reasons are we can rest assured that:
$u/2$ is just as positive as $u$ is, and
It is smaller than $u$. Both of these will be important.
Now from the limit definition, and the continuity of $f$, we get our fat, grubby hands on a $\delta$ that guarantees that as long as |x - c| < \delta, we can be confident that |f(x) - u| < \frac u 2. However, what we actually want to do is prove $f$ is positive on such a range of $x$'s. So the absolute value signs aren't fair: that automatically positifies everything. So we take it away: recall that if t < 0 < s, then $|t| = -t > 0 > -s$. This lets us re-write the inequality |f(x) - u| < \frac u 2 as the double inequality:
-\frac u 2 < f(x) - u < \frac u 2
where what we have is honest-to-midwestern-goodness real numbers, no more hiding behind those absolute value signs, no sir. Now we can add $u$ to all three terms, without fear of messing up the inequalities:
-\frac u 2 + u < f(x) - u + u < \frac u 2 + u \iff\frac u 2 < f(x) < 3\frac u 2
The inequality on the right-hand isn't very interesting to us, knowing that f(x) < 3u/2 doesn't reveal anything about whether or not $f(x)$ is positive. But the one on the left is pure gold: it tells us that on the entire real interval $(x-\delta,x+\delta)$, that u/2 < f(x). And $u/2$ is positive, so $f(x)$ is even "more positive". And on that happy note, I bid you adieu.