Determining a value $k$ that gives a size = 0.05 test. A single observation $X$ from a normal distribution with mean $\mu$ and $\sigma^2$=1 is used to test $$H_0 : \mu = 1 \ \ \ \text{vs} \ \ \ H_1 : \mu \lt 1 $$
using the critical region $C = {{x : x \lt k}}$
Determine the value of k that gives a size 0.05 test.
My attempt:
The size of the test = significance level = $\alpha$ = 0.05
From my notes, I am told $\alpha = \pi(H_0)$, where $\pi$ signifies the power function.
The only thing I could think of that might link these together is the z-score formula. $$ Z_{0.05} = 1.65 $$ $$ 1.65 = \frac{\bar x - 1}{1} $$
However this doesn't make sense to me, as when I solve for $\bar x$ I get 2.65, which is greater than $\mu$, and this region should be less than $\mu$
Am I supposed to derive the power function for $\mu$ myself?

You have most of the pieces. Let's put them together: You want to reject $H_0: \mu = 1$ against
$H_a: \mu < 1$ at the 5% level when $X < k.$
Thus, under $H_0$ you want $P(X < k) = 0.05.$
Putting this into a form so that you can use printed standard normal tables, we have
$$P\left(\frac{X - \mu_0}{\sigma} = X - 1 < k - 1\right) = P(Z < k-1) = .05,$$
where $\mu_0 = 1,\, \sigma = 1,$ and $Z \sim \mathsf{Norm}(0,1),$ that is, standard normal.
From the table you have $P(Z < - 1.645) = 0.05.$ So $k - 1 = -1.645,$
and finally $k = 1 - 1.654 = -0.645.$
In this simple case, rejecting $H_0$ when $X < -.0645$ is the same as
rejecting when $Z < -1.645.$ (The latter inequality is the 'usual' way 
to state the rejection criterion.)
