I'm looking for feedback on my work, since we've only just started talking about continuity and differentiability and I'm not 100% confident in my knowledge.. I'd love to know if there are any inaccuracies or general tips for improvement. :)
We know $f$ is continuous on $[0,1]$,
then by Weierstrass's Theorem there exist $a ∈ [0,1]$ such that $f(a) ≤ f(x)$, for every $x ∈ [0,1]$.
Consider the fucntion $g: [0,f(a)]➜[0,f(a)]$, $\ g(x) = x$.
Note that by the algebra of continuous functions $g$ is continuous on $[0,1]$.
Since $f(x)>0$ for every $x ∈ [0,1]$, and since $g(0) = 0$,$ \ g(f(a))=f(a)$, then we have
$g(0) < \frac{f\left(a\right)}{2} < g(f(a))$.
Denote $\frac{f\left(a\right)}{2} = ϵ$.
Then we proved the existance of an ϵ > 0 such that f(x) > ϵ for every x ∈ [0,1].
I was originally planning to use the intermediate value theorem after finding the element between the two values of g, but then I noticed that I don't actually need to and I've already found such an element.. Is there anything I can improve?
I see no reason for introducing that function $g$. You defined $\varepsilon$ as $\frac{f(a)}2$. Then you can simply say that, for each $x\in[0,1]$,\begin{align}f(x)&\geqslant f(a)\\&>\frac{f(a)}2\text{ (since $f(a)>0$)}\\&=\varepsilon.\end{align}