1. Lemma Statement and Proof:
We start with the lemma: If $f$ is a polynomial with integer coefficients and $\deg f < p-1$, then
$$f(0) + f(1) + \cdots + f(p-1) \equiv 0 \pmod{p}.$$
To prove this lemma, we can reduce it to the case of monomials. Consider a monomial $f(x) = x^k$ where $k < p-1$. By Fermat's Little Theorem, we know that for any integer $a$,
$$a^p \equiv a \pmod{p}.$$
This implies that for $k < p-1$,
$$\sum_{i=0}^{p-1} i^k \equiv 0 \pmod{p}.$$
This is because the sum of the $k$-th powers of the first $p-1$ integers modulo $p$ is zero for $k < p-1$. Therefore, the lemma holds for any polynomial $f$ of degree less than $p-1$.

2. Application of the Lemma:
Given the polynomial $f$ with $f(0) = 0$ and $f(1) = 1$, and $f(n) \equiv 0$ or $1 \pmod{p}$ for every integer $n$, we need to show that $\deg f \geq p-1$.

3. **Summing $f(i)$ for $i = 0, 1, \ldots, p-1$:**
Since $f(i) \equiv 0$ or $1 \pmod{p}$ for $i = 0, 1, \ldots, p-1$, the sum
$$f(0) + f(1) + \cdots + f(p-1)$$
is an integer between $0$ and $p-1$.

4. **Contradiction if $\deg f < p-1$:**
If $\deg f < p-1$, by the lemma, we have
$$f(0) + f(1) + \cdots + f(p-1) \equiv 0 \pmod{p}.$$
However, since each $f(i)$ is either $0$ or $1$, the sum $f(0) + f(1) + \cdots + f(p-1)$ is an integer between $0$ and $p-1$. The only way this sum can be congruent to $0 \pmod{p}$ is if the sum is exactly $0$, which is impossible because $f(1) = 1$.

5. Conclusion:
Therefore, the assumption that $\deg f < p-1$ leads to a contradiction. Hence, we must have
$$\deg f \geq p-1.$$

$\blacksquare$