Let $I_{n, d}$ be the integral $\int_{d - 1}^d\binom x n dx$. Then your $I_n$ is just $\sum_{d = 1}^n I_{n, d}$.
We are going to treat the terms $I_{n, d}$ in several cases.
As a first observation, note that on the interval $[d - 1, d)$, we have:
\begin{eqnarray}
n!\left|\binom x n\right| &=& x \cdot (x - 1)\cdots (x - (d - 1)) \cdot (d - x) \cdot (d + 1 - x) \cdots (n - 1 - x)\\
&\leq& d \cdot (d - 1) \cdots 1 \cdot 1\cdot 2 \cdots (n - d)\\
&=& d! (n - d)!
\end{eqnarray}
which gives $\left|\binom x n\right| \leq \binom n d ^{-1}$.
Now for our $I_{n, d}$, we have the following cases.
Case 1: $d = 1$ or $d = n - 1$.
For these two $d$, we have $\left|\binom x n\right| \leq n^{-1}$, hence $I_{d, n} \leq n^{-1}$. Taking limit yields $\lim_{n\rightarrow \infty} I_{n, 1} + I_{n, n - 1} = 0$.
Case 2: $2\leq d \leq n - 2$.
In this case, we have $\left|\binom x n\right| \leq \frac 2{n(n - 1)}$ for sufficiently large $n$ (e.g. $n \geq 4$). Therefore we can bound the sum:
$$\sum_{d = 2}^{n - 2} \left|I_{n, d}\right|\leq (n - 3)\cdot \frac 2{n(n - 1)}.$$ Taking limit again gives $\lim_{n \rightarrow\infty} \sum_{d = 2}^{n - 2} I_{n, d} = 0$.
Case 3: $d = n$.
This is the only remaining case. We write $J_n$ for $I_{n, n}$ and rewrite the integral as $$ J_n = \int_{n - 1}^n\binom x n dx = \int_0^1 \frac{x \cdot (x + 1) \cdots (x + n - 1)}{n!}dx.$$
Write $g_n(x) = \frac{x \cdot (x + 1) \cdots (x + n - 1)}{n!}$. Obviously we have $0\leq g_n(x) \leq 1$ for all $x\in[0, 1]$.
Furthermore, it is a simple exercise, based on the divergence of the harmonic series, that $\lim_{n\rightarrow \infty} g_n(x) = 0$ for any $x\in[0, 1)$.
Now the magic happens by applying the dominated convergence theorem. It tells us that the limit $\lim_{n\rightarrow\infty} J_n$ is $0$.
Combining all three cases, we get $\lim_{n\rightarrow\infty} I_n = 0$.