Before trying to find a UMP test, one needs to first check if there exists one. To do this one needs to find the likelihood ratio function
$$l(x)=f_{\theta_1}(x)/f_{\theta_0}(x)$$
This function must be monotone non-decreasing in $x$ for every $\theta_1\geq \theta_0$. In the given question $\theta_1=2$, and the density function is $$f_{2}(x)=2x.$$ Similarly for $\theta_0\in[1/2,1]$, $$f_{\theta_0}(x)=\theta_0x^{\theta_0-1}$$ Hence, the likelihood ratio function is
$$l_{\theta_0}(x)=\frac{2x}{\theta_0x^{\theta_0-1}}=\frac{2}{\theta_0}x^{2-\theta_0}$$
Since this function is increasing in $x$ for all $\theta_0\in[1/2,1]$, there exists a UMP test of level $\alpha$.
By definition of UMP test, the significance level $\alpha$ is the expected value of the decision rule (which is the likelihood ratio test with a certain threshold $\lambda$), for which the false alarm probability lies below $\alpha$, for every $\theta_0$
$$\alpha=\sup_{\theta_0}\int_{\{x:l_{\theta_0}(x)>\lambda\}}f_{\theta_0}(x)\mathrm{d}x=\sup_{\theta_0}\int_{\{x:l_{\theta_0}(x)>\lambda\}}\theta_0x^{\theta_0-1}\mathrm{d}x$$
Now, we have a nice simplification (Why?) $${\{x:l_{\theta_0}(x)>\lambda\}}\equiv {\{x:x>\lambda^{'}\}}$$
Hence
$$\alpha=\sup_{\theta_0}\int_{\{x:l_{\theta_0}(x)>\lambda\}}\theta_0x^{\theta_0-1}\mathrm{d}x=\sup_{\theta_0}\int_{\lambda^{'}}^1\theta_0x^{\theta_0-1}\mathrm{d}x=\sup_{\theta_0}1-{\lambda^{'}}^{\theta_0}=0.05$$
It is known that $\lambda^{'}\in[0,1]$ and $\theta_0\in[1/2,1]$. Now what value of $\theta_0$ maximizes $1-{\lambda^{'}}^{\theta_0}$ or similarly minimizes ${\lambda^{'}}^{\theta_0}$?
The UMP test is then $$\phi(x)=\begin{cases}1,\quad x>\lambda^{'}\\0,\quad x\leq \lambda^{'}\end{cases}$$
All was well done up to the line "then $c=1−\alpha$ will produce a $\alpha$ level test."
The test that you have constructed is already the most powerful $\alpha$-level test according to Neyman-Pearson Lemma.
It seems to me that you do not quite understand the definition. There are many tests for these two hypotheses whose level equal to or does not exceed $\alpha$.
For example, the test with a critical region $X<\alpha$ also has an $\alpha$ level.
The test with a critical region $X\in[0,\frac{\alpha}{2})\cup(1-\frac{\alpha}2,1]$ also has an $\alpha$ level. And so on.
But among all of them, the likelihood ratio criterion of $\alpha$ level has the greatest power. This is the statement of the Lemma.
So to construct MP $\alpha$-level test you need to constuct a likelihood ratio test and equate its level to $\alpha$, which exactly you did.
You can in addition calculate its power $\beta=1-(1-\alpha)^2$ but it is not needed to do smth with it. You can admire this value. You can say that here it is - the biggest power of the test of level $\alpha$, more than which no test of level $\alpha$ can ever have. But hardly anything else.
Best Answer
The Neyman Pearson Lemma allows you to find the MP test at level $\alpha$ in the following way:
Then you get the likelihood ratio as,$$\left(\dfrac{\theta_1}{\theta_0}\right)^n\prod_{i=1}^nX_i^{\theta_1-\theta_0}\geq K$$
Now you are given $\theta_1>\theta_0$. So in particular, taking log on both sides, and noting that $\theta_1/\theta_0$ and $\theta_1-\theta_0$ are afterall constants, you essentially get that for some $K^*$ you have $$\log\left(\prod_{i=1}^nX_i\right)\geq K^*$$.
Essentially, you have then by taking exponential function on both sides,
$$\prod_{i=1}^n X_i\geq C$$ for some $C\geq0$.
So rejection indeed occurs at large values of your sufficient statistic $T=\prod_i X_i$.