Show that if $\text{card}(X) \geq 2$, there is a topology on $X$ that is $T_0$ but not $T_1$
Consider the point set $\{a,b\}$ where $b>a$. Then, the topological space of this set is $T_0$ as there is an open neighborhood of one that does not contain the other. But this is not $T_1$ as $b$ won't be closed. By induction, the following holds for $\text{card}(X) \geq 2$. Is this okay?
No induction needed. Let $p$ and $q$ be distinct points of $X$.
Define $\mathcal{T} = \{A \subseteq X: p \in A\} \cup \{\emptyset\}$, which is a topology (the included point topology).
Then $(X,\mathcal{T})$ is $T_0$ (if $x \neq y$ then one of them is not $p$ (say $x$) and then $X\setminus\{x\}$ is open and contains $y$ and not $x$), but not $T_1$ as $\{p\}$ is not closed, as $X\setminus \{p\}$ is not open (it's non-empty because of $q$ and does not contain $p$, so it's not in $\mathcal{T}$).
As an alternative, put a linear order on $X$ (uses axiom of choice) and we have two points $p < q$ WLOG.
Define the topology (upper topology) by $\mathcal{T}_u=\{\emptyset,X\} \cup\{(x,\infty): x \in X\}$, where $(x,\infty) = \{p \in X: x < p\}$. This is $T_0$ because if $x \neq y$, then $x < y$ (WLOG) and then $y \in (x,\infty)$ and $x$ is not in that set. It is not $T_1$, e.g. because $p$ will always be in the closure of $\{q\}$ when $p < q$ (as chosen above).