The question goes as follows:
Let $(X, \mathcal{T})$ be a topological space. Define a subset of $X^2 = X \times X$ called the diagonal by $$ \Delta := \{(x, x) \in X^2: x \in X\} $$ Show that $(X, \mathcal{T})$ is homeomorphic to $\Delta$ with its subspace topology inherited from the topology on $X^2$
And here is my solution:
To show that $(X, \mathcal{T})$ is homeomorphic to $\Delta$ with its subspace topology we need to find a homeomorphism between the two. For this let's define the map $\phi: X \to \Delta$ as follows: $$ \phi(x) = (x, x) $$ It is clear that $\phi$ is a bijection. We need to show that both $\phi$ and $\phi^{-1}$ are continuous. For $\phi$ to be continuous, We need to show that the preimage of any open set in $\Delta$ under $\phi$ is open in $X$.
Let $U \subseteq \Delta$ be an open set in the subspace topology on $\Delta$. Then, there exists an open set $V$ in $X^2$ such that $U = \Delta \cap V$. Since $U$ is open, for every $(x, x) \in U$, there exists a neighborhood $W$ of $(x, x)$ in $X^2$ such that $W \cap \Delta \subseteq U$. Now, consider the preimage $\phi^{-1}(U)$ in $X$. We have: \begin{align*} \phi^{-1}(U) &= \phi^{-1}(\Delta \cap V) \\ &= \{x \in X: \phi(x) \in \Delta \cap V\} \\ &= \{x \in X: \phi(x) \in U\} \\ &= \{x \in X: (x, x) \in U\} \\ &= \{x \in X: (x, x) \in \Delta \cap V\} \\ &= \{x \in X: (x, x) \in W \cap \Delta\} \\ &= \{x \in X: (x, x) \in W\} \\ &= W_X, \end{align*} where $W_X$ is the projection of $W$ onto the $X$-coordinate. Since $W$ is an open set in $X^2$, $W_X$ is open in $X$. Thus, $\phi^{-1}(U)$ is open in $X$. This shows that $\phi$ is continuous.
$\phi^{-1}$ is continuous: We need to show that the preimage of any open set in $X$ under $\phi^{-1}$ is open in $\Delta$.
Let $U \subseteq X$ be an open set in $X$, and consider the preimage $\phi^{-1}(U)$ in $\Delta$. We have: \begin{align*} \phi^{-1}(U) &= \{x \in X: \phi(x) \in U\} \\ &= \{x \in X: (x, x) \in U\} \\ &= \{(x, x) \in \Delta: x \in U\} \\ &= \Delta \cap (U \times U). \end{align*} Since $U \times U$ is open in $X^2$, its intersection with $\Delta$ is open in $\Delta$ (inherited subspace topology). Therefore, $\phi^{-1}(U)$ is open in $\Delta$. This shows that $\phi^{-1}$ is continuous.
Since $\phi$ is a bijection and both $\phi$ and $\phi^{-1}$ are continuous, we can conclude that $(X, \mathcal{T})$ is homeomorphic to $\Delta$ with its subspace topology.
Is my solution correct? Is there any simpler solution to this?
Your solution is not quite right. There are two easily fixed mistakes in showing $\phi$ is continuous.
Specifically, it isn't true in general that: $$\{x\in X:(x,x)\in\Delta\cap V\}=\{x\in X:(x,x)\in W\cap\Delta\}$$After all, you took $W$ to be any subneighbourhood of $(x,x)$ in $V$ (where $(x,x)$ is quite arbitrary); $W\subsetneq V$ may be true. There's also no point in talking about $W$ in the way you did. If you just want $W$ to be any subneighbourhood, you can take $V$ itself! $V$ itself is open!
It is true that the projections $\pi_{1,2}:X\times X\to X$ are open maps. But you make two mistakes: firstly, "$W_X$" is ambiguous: both coordinates are the "$X$"-coordinate. Secondly, it isn't true in general that: $$\{x\in X:(x,x)\in W\}=W_X$$For either interpretation ($\pi_1(W)$ or $\pi_2(W)$) of "$W_X$". Consider $X=\Bbb R$ and $W=(0,1)\times(0,1/2)$. This is a neighbourhood of some elements of the diagaonl $\Delta$, however $\pi_1W=(0,1)$ and $\pi_2W=(0,1/2)$ are not the same and e.g. $2/3\in\pi_1W$ but $2/3\notin\pi_2W$, so it's not true that $x\in W_X\implies(x,x)\in W$.
The fix is to make a specific choice of subneighbourhood of $V$. I don't just want $W\subseteq V$ to be any old open set this time: fix $(x,x)\in V$ and choose an open neighbourhood $W=W_1\times W_2\subseteq V$ of $(x,x)$ where both $W_1,W_2$ are open in $X$. This is possible by definition of the product topology. Note: it's generally only advantageous to look at subneighbourhoods $W$ if you can choose them to be 'special' somehow, e.g. choosing them from a basis. I leave it as an exercise to consider $W_1\cap W_2$ and conclude $\phi$ is continuous.
In more detail, here's how we can see $\phi$ is continuous. For ease of notation, I consider $\phi$ to be a map $X\to X\times X$; if this is continuous, then the map $X\to\Delta$ is also continuous.
We let $x\in X$ be arbitrary, $V\ni\phi(x)=(x,x)$ any neighbourhood. There are open neighbourhoods $W_1,W_2$ of $x$ with $W:=W_1\times W_2\subseteq V$ by definition of the product topology. Consider $U:=W_1\cap W_2$. This is an open neighbourhood of $x$.
$$\phi(U)=\{(y,y):y\in U\}=\{(y,y):y\in W_1\wedge y\in W_2\}\subseteq W_1\times W_2\subseteq V$$
So, for any neighbourhood $V$ of $\phi(x)$ there is a neighbourhood $U$ of $x$ with $\phi(U)\subseteq V$. That is, $\phi$ is continuous "at $x$". Since $x\in X$ was arbitrary, $\phi$ is continuous.
Oh, and you should not write $\phi^{-1}(U)$ when you're trying to conclude $\phi^{-1}$ is continuous. You really meant, $(\phi^{-1})^{-1}(U)$ i.e. $\phi(U)$. It's simpler and equivalent to think about showing that $\phi$ is an open map.
A much, much quicker way to conclude $\phi$ is continuous: the map $x\mapsto(x,x),X\to X\times X$ is continuous because it is the product pairing of the continuous identity function with itself (re: universal property of the product space). So, the 'corestriction' of this map to $\phi:X\to\Delta$ will also be continuous.