Is one able to say that the average weight of candy bars is at least 18.2g?
$n = 10$
$\bar{x} = 18.57$
$\sigma = 0.2$
level of significance $\alpha = 5.71\%$
$$H_{0} = \mu \leq \mu_{0}$$ $$H_{1} = \mu > \mu_{0}$$
(I'm not sure whether this is correct. I've heard that the thing you're trying to prove should always be $H_{1}$. The thing we're trying to prove is $\mu \geq \mu_{0}$, but then i heard that the equal sign should never be in $H_{1}$)
Assuming my $H_{0}$ and $H_{1}$ are correct:
Critical value of $\alpha$ : $-1,579$
$$ z=\frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}}=\frac{18.57-18.2}{\frac{0.2}{\sqrt{10}}}= 5.85$$
Since z-value > critical value, reject $H_{0}$
First, note that the null hypothesis or $H_0$ is the statement that we try to reject. If we succeed in rejecting $H_0$ then we can accept the alternative hypothesis, or $H_1$. So we always formulate our work such that we can (maybe) reject the case that we are not trying to show!
A small but important point is that we do not assume that $H_0$ and $H_1$ are both correct. Rather, we start by assuming that $H_0$ is correct, and we will see if it leads us to a conclusion that is probable or not probable (and our criteria for deciding what probabilities to accept is the level of significance). Edit: I may have misunderstood your sentence there. If you meant to assume that you have formulated $H_0$ and $H_1$ correctly, then I agree with you. (Thanks @5201314 for pointing it out)
Now, your calculation of z-value is correct. But there is a problem with the critical value that you identified. I see you have checked the tables (or used a software) to find the critical value corresponding to $\alpha = 5.71\%$. This is a mistake in this exercise. Here, you should find the critical value corresponding to $100\%-\alpha$ (or $1-\alpha$ if it is written as a proportion). This is because of the $H_0$ of this problem: we are assuming that the mean is less than or equal to $\mu_0$. You might say that the improbable values that violate our assumption are then very large values. The figure below shows the idea.
Finally, your conclusion is correct, because it just happens that the z-value is still larger than the correct critical value, but in an exam, your answer would probably lose marks.