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$G\times X$ is a product topology iff $X$ is $T_0$

Let $X$ be a topological space.
I would like to show that the topology on $G\times X$ given by the subbase
$$\big\{ U\times V: U \text{ is open in } G\big\}$$
is equal to the product topology if and only if $X$ is $T_0$.
I tried using the subbase for $X$, but I couldn’t make it. I really have no clue how to start this one. Please help me!
Thanks.

No, the topology is not equal to the product topology if $X$ is not $T_0$, e.g. the two-point space $\{0,1\}$ is $T_0$, but $\mathbb R$ with the discrete topology is not.
A few hints:

If $X$ is $T_0$, consider $X\times\{0\}$ and its subbase;
If $X$ is not $T_0$, consider the open sets $U=\{0\}\times X$, $V=X\times\{1\}$.

Determine whether user roles are present in role enumeration

I have this code for checking whether the user has all the roles assigned and displaying that as a message. I’m using both the default and custom role enumeration. Now I need a functionality where I need to ensure that the user also has roles assigned to the role they have access to. How can I do that in a simple manner?
// Constructor
public function __construct($user = null)
{
parent::__construct();

$this->user = $user;
}

// Check if user has the specified roles
public function hasRoles($roles)
{
$user = (null!== $this->user)? $this->user : $this->getUser();

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