Is the sum and difference of two irrationals always irrational?

The short answer to your question is that is not necessarily true. For instance, $\sqrt{2}, \sqrt{2}-1,1-\sqrt{2}$ are all irrational but $$\sqrt{2} + (1-\sqrt{2}) = 1 \in \mathbb{Q}$$ and $$\sqrt{2} - (\sqrt{2}-1) = 1 \in \mathbb{Q}$$

However, it is worth noting that if $x$ and $y$ are irrational, then either $x+y$ or $x-y$ is irrational i.e. $x+y$ and $x-y$ cannot be both rationals. The proof for this is given below.

Proof

If both $x+y$ and $x-y$ are rational, then we have that $x+y = \dfrac{p_1}{q_1}$ and $x-y = \dfrac{p_2}{q_2}$, where $p_1,p_2 \in \mathbb{Z}$ and $q_1,q_2 \in \mathbb{Z} \backslash \{0\}$.

Hence, $x = \dfrac{\dfrac{p_1}{q_1} + \dfrac{p_2}{q_2}}{2} = \dfrac{p_1q_2 + p_2q_1}{2q_1q_2}$ and $y = \dfrac{\dfrac{p_1}{q_1} - \dfrac{p_2}{q_2}}{2} = \dfrac{p_1q_2 - p_2q_1}{2q_1q_2}$.

Now $p_1q_2 + p_2q_1, p_1q_2 - p_2q_1 \in \mathbb{Z}$, whereas $2q_1q_2 \in \mathbb{Z} \backslash \{0\}$.

This contradicts the fact that $x$ and $y$ are irrationals. Hence, if $x$ and $y$ are irrational then either $x+y$ is irrational or $x-y$ is irrational.

Below are some statements worth knowing.

1 Sum of two rationals is always a rational.

Proof: Let the two rationals be $\dfrac{p_1}{q_1}$ and $\dfrac{p_2}{q_2}$, where $p_1,p_2 \in \mathbb{Z}$ and $q_1,q_2 \in \mathbb{Z} \backslash{0}$. Then $$\dfrac{p_1}{q_1} + \dfrac{p_2}{q_2} = \dfrac{p_1q_2 + p_2q_1}{q_1q_2}$$
where $p_1q_2 + p_2q_1 \in \mathbb{Z}$ and $q_1 q_2 \in \mathbb{Z} \backslash \{0\}$. Hence, the sum is again a rational.

2 Sum of a rational and an irrational is always irrational.

Proof: Let the rational number be of the form $\dfrac{p}{q}$, where $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$ while the irrational number be $r$. If $r + \dfrac{p}{q}$ is a rational, then we have that $r + \dfrac{p}{q} = \dfrac{a}{b}$ for some $a \in \mathbb{Z}$ and $b \in \mathbb{Z} \backslash \{0\}$. This means that $r = \dfrac{a}{b} - \dfrac{p}{q} = \dfrac{aq-bp}{bq}$ where $aq-bp \in \mathbb{Z}$ and $bq \in \mathbb{Z} \backslash \{0\}$. This contradicts the fact that $r$ is irrational. Hence, our assumption that $r + \dfrac{p}{q}$ is a rational is false. Hence, $r + \dfrac{p}{q}$ is a irrational.

3 Sum of two irrationals can be rational or irrational.

Example for sum of two irrationals being irrational

$\sqrt{2}$ is irrational. $\sqrt{2} + \sqrt{2} = 2 \sqrt{2}$ which is again irrational.

Example for sum of two irrationals being rational

$\sqrt{2}$ and $1-\sqrt{2}$ are irrational. (Note that $1-\sqrt{2}$ is irrational from the second statement.) But, $\sqrt{2} + (1-\sqrt{2}) = 1$ which is rational.