$f:[a, b] \rightarrow \mathbb{R}$ continous and $f(x) > 0$ for all $x$, then there exist $L > 0$ such that $f(x) \geq L$ for all $x$ $f:[a, b] \rightarrow \mathbb{R}$ continous and $f(x) > 0$ for all $x$, then there exist $L > 0$  such that $f(x) \geq L$ for all $x$.
My approach is:
$f([a, b])$ is bounded below by $0$, so we can consider $\alpha = \inf f([a, b])$. $0$ is a lower bound so $\alpha \geq 0$. Suppose $\alpha = 0$, because $f$ is a continuous function on a compact set it attains it's minimum, so in this case there must be an $x \in [a, b]$ such that $f(x) = \alpha = 0$, but this contradicts that $f(x) > 0$, so it must be the case that $f(x) \geq L$ with $L > 0$ for all $x$.
Is this correct?
 A: If you already know that a continuous function on a compact set attains its minimum, you can significantly simplify your proof:

Let $x_m\in [a,b]$ be the value at which $f$ attains its minimum. Then
set $L=f(x_m)$. By assumption, $L=f(x_m)>0$, and by the definition of
minimum, $f(x)\geq L$ for all $x$.

No proof by contradiction necessary.
A: Yes, it's correct. The core observation is, of course, that $f$ must actually attain its minimum.
Some nit-picks. There is no reason to use $\alpha$, just use $L$ from the start. Also, I am a huge fan of not using more contradiction than necessary. There is no need to assume $\alpha=0$ (or really $L=0$). Finally, you use $x$ for two different things (both as an arbitrary point in the interval and as the specific point where $f$ has its minimum), which is not ideal.
So my preferred version of your proof would be

$f([a, b])$ is bounded below by $0$, so we can consider $L = \inf f([a, b])$. $0$ is a lower bound so $L\geq 0$. Because $f$ is a continuous function on a compact set it attains its minimum, so in this case there must be an $x_0 \in [a, b]$ such that $f(x_0) = L$. But since $f(x)>0$ for all $x\in [a,b]$, in particular $L=f(x_0)>0$. So it must be the case that $f(x) \geq L$ with $L > 0$ for all $x$ in $[a,b]$.

