Properties of Pushout suppose we have a pushout square in $\mathrm{Top}$:
\begin{align*}
 \require{AMScd}
 \begin{CD}
  X_0 @>{\mu_1}>> X_1\\
  @V{\mu_2}VV @VV{\alpha_1}V \\
  X_2 @>>{\alpha_2}> X
 \end{CD}
\end{align*}
I already know that $X= X_1 \sqcup X_2/\sim$ up to homeomorphism and 
I already showed (with help) that the following 3 statements hold:
1. $X = \alpha_1(X_1) \cup \alpha_2(X_2)$
2. $\alpha_1(X_1) \cap \alpha_2(X_2) = \alpha_1(\mu_1(X_0)) = \alpha_2(\mu_2(X_0))$
3. If $\mu_1$ is injective then $\alpha_2$ is injective too.
Now, I would like to show that
4. If $\mu_1$ is injective and $\mu_1(X_0) \subseteq X_1$ is closed then $\alpha_2(X_2) \subseteq X$ is closed too. 
Unfortunately, I have no idea how to start. Could somehow help me out and give me a hint?
Here, just some random thoughts I have (not very constructive though) 
We know that $\mu_1$ is injective and by 3. it follows that also $\alpha_2$ is injective. Thus, it holds that
\begin{align*}
\mu_1^{-1}(\mu_1(X_0)) = X_0 \\
\alpha_2^{-1}(\alpha_2(X_2)) = X_2
\end{align*}
Since, $\mu_1$ and $\alpha_2$ are continuous, we have that $X_0$ and $X_2$ are closed sets... And now, I have no idea...  
Cheers,
quizzle123

Note that $\mu_1^{-1}(\mu_1(X_0))=X_0$ is almost a tautology since clearly $X_0$ is the set of points in $X_0$ which are mapped to $\mu_1(X_0)$. Maybe you meant $\mu_1^{-1}(\mu_1(A))=A$ for every $A\subseteq X_0$, and this is indeed the injectivity of $\mu_1$.
Note that $X$ is the quotient space $(X_2⊔X_1)/(μ_2(x)∼μ_1(x))$. Let $q:X_2⊔X_1\to X$ denote the quotient map. A set $B$ in $X$ is closed if and only if $q^{-1}(B)$ is closed. So in order to show that $\alpha_2(X_2)$ is closed you have to show that $q^{-1}(α_2(X_2))$ is closed in $X_2⊔X_1$. Can you prove that $q^{-1}(α_2(X_2))=X_2⊔\mu_1(X_0)$ ?
