To construct the UMP test, we have to construct the corresponding MP test. Hence, the LR is given by
\begin{align}
\frac{L_{1}(X_1,..,X_n|\lambda_1)}{L_{1}(X_1,..,X_n|\lambda_0)} = \frac{\prod_{i=1}^n \frac{e^{-\lambda_1} \lambda_1^{x_i}}{x_i!}}{\prod_{i=1}^n \frac{e^{-\lambda_0} \lambda_0^{x_i}}{x_i!}} &= \exp\{n(\lambda_0 - \lambda_1)\}\left(\frac{\lambda_1}{\lambda_0}\right)^{\sum_i^nx_i}\\
&= \exp\{n(\lambda_0 - \lambda_1)\}\left(\frac{\lambda_1}{\lambda_0}\right)^{n\bar{x}_n} > c\,.
\end{align}
This statistic depends on the distribution of $\bar{X}_n$, hence for large enough $n$ we can use the CLT to approximate the rejection region. Note that the MP is $\Psi(\mathrm{X}) = \mathcal{I}\left( \bar{X}_n >c' \right)$.
\begin{align}
\alpha = \mathbb{E}_{\lambda_0}\Psi(\mathrm{X}) &= \mathbb{P}_{\lambda_0}\left( \bar{X}_n >c' \right)\\
&a\approx 1-\phi\left(\frac{c'-\lambda_0}{\sqrt{\lambda_0/n}} \right)\\
&c' = \lambda_0 + Z_{1-\alpha}\sqrt{\lambda_0/n}.
\end{align}
For $\lambda_0 = 1$,
$$
c' = 1 + Z_{1-\alpha}\sqrt{1/n}\,.
$$
- Power function for the parametric space $\Lambda = \mathbb{R}^+$.
\begin{align}
\pi(\Psi(\mathrm{X})|\Lambda) &= \mathbb{E}_{\Lambda}\Psi(\mathrm{X}) = \mathbb{P}_{\Lambda}(\bar{X}_n>c')\\
&=1-\phi\left( \frac{c'-\lambda}{\sqrt{\lambda/n}} \right),& \forall \lambda \in \Lambda.
\end{align}
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 $$\phi(x)=\begin{cases}1,\quad x>\lambda^{'}\\0,\quad x\leq \lambda^{'}\end{cases}$$
Best Answer
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.