"What is the probability of drawing either an ace or a diamond in a single draw from a fair deck of cards?" For this question would I need to subtract the ace of diamonds in my calculation because it is both a diamond and an ace and I would not want to calculate this twice. Here is what I did:
$$\begin{align} \mathbb{P}(\text{ ace or diamond }) &= \frac{\# \text{ ace or diamond} }{\# \text{ total}}\\ & = \frac{4 + 13 - 1}{52} \\ &\approx 0.30769 \end{align}$$
$-1$ was done to remove common repeated card "ace of diamond".
You have already received the answer you expected in comments, but here is a detailed answer of why it is the good way to compute this probability.
Suppose that you have a finite set $\Omega$ (here, $\Omega$ is the set of all $52$ cards in your deck). You consider here that the events are $\mathcal{P}(\Omega)$, i.e. the so-called power set of $\Omega$. The power set of $\Omega$ is defined as the set of all subsets of $\Omega$. For example, $\mathcal{P}(\Omega)$ contains the set $\{\text{Ace diamond },\text{ Jack club },\text{ 7 club },\text{ 8 heart }\}$.
Actually, $\mathcal{P}(\Omega)$ contains all the "events" on $\Omega$. For example, the event "ace or diamond" is $$\begin{align} &\{\text{Cards }\in\Omega\mid \text{Cards are aces or diamonds}\}\\ &=\{\text{ Ace diamond },\text{ Ace heart }, \text{ Ace club },\text{ Ace spade }, \color{red}{2,3,4,5,6,7,8,9,10,\text{ jack }, \text{ queen },\text{ king } \text{ diamonds }}\} \end{align}$$ which is clearly a subset of $\Omega$, so that it belongs to $\mathcal{P}(\Omega)$.
Equiprobability implies that the probability that an event $A\in\mathcal{P}(\Omega)$ occurs is equal to the number of elements in $A$ divided by the number of elements in $\Omega$. We shall denote $\vert A\vert$ the number of elements in a set $A$. We shall denote $A\cup B$ the union of $A$ and $B$, which is the set of the elements that are in $A$ or in $B$ or in both, while $A\cap B$ denotes the intersection of $A$ and $B$, which is the set of elements that are in $A$ AND in $B$.
A famous formula (known as the inclusion-exclusion principle) is the following:
$$\vert A\cup B\vert = \vert A\vert + \vert B\vert - \vert A\cap B\vert\tag{1}$$
which can be read "The number of elements that are in $A$ or in $B$ or in both is equal to the number of elements in $A$ plus the number of elements in $B$, minus the elements in $A$ and $B$". Why is that intuitively true? Because if there are elements in $A$ and in $B$, when we make $\vert A\vert +\vert B\vert$, we double count the elements that are both in $A$ and $B$, so that we have to take out one time these elements, which are by definition $A\cap B$.
Let's write $A=\{\text{Cards }\in\Omega\mid \text{Cards are aces or diamonds}\}$. Now, you have (because of equiprobability, which is the main argument here):
$$ \mathbb{P}[A] = \frac{|A|}{|\Omega|} $$
Here, $\vert \Omega\vert = 52$. But what is $\vert A\vert$ in your case? We can rewrite:
$$\begin{align} &\{\text{Cards }\in\Omega\mid \text{Cards are aces or diamonds}\}\\ &=\{\text{Cards }\in\Omega\mid \text{Cards are aces}\}\,\cup\,\{\text{Cards }\in\Omega\mid \text{Cards are diamonds}\} \\ \end{align}$$
So, if we use formula $(1)$ to compute $\vert A\vert$:
$$\begin{align} \vert A\vert &=\left\vert\{\text{Cards }\in\Omega\mid \text{Cards are aces or diamonds}\}\right\vert\\ &=\left\vert \{\text{Cards }\in\Omega\mid \text{Cards are aces}\}\,\cup\,\{\text{Cards }\in\Omega\mid \text{Cards are diamonds}\}\right\vert\\ &=\color{blue}{\left\vert \{\text{Cards }\in\Omega\mid \text{Cards are aces}\}\right\vert}+\color{green}{\left\vert\{\text{Cards }\in\Omega\mid \text{Cards are diamonds}\}\right\vert} -\color{red}{\left\vert\{\text{Cards }\in\Omega\mid \text{Cards are aces AND diamonds}\}\right\vert}\\ &= \color{blue}{4} + \color{green}{13} - \color{red}{1} \end{align}$$ so that $$\begin{align} \mathbb{P}[A] &= \mathbb{P}\left[\{\text{Cards }\in\Omega\mid \text{Cards are aces or diamonds}\}\right]\\ &=\frac{\left\vert\{\text{Cards }\in\Omega\mid \text{Cards are aces or diamonds}\}\right\vert}{\vert\Omega\vert}\\ &=\frac{4+13-1}{52}\\ &=\frac{4}{13}\\ &\approx 0.30769 \end{align}$$