My solution uses two main ingredients.
1) Frobenius theorem ensures complete integrability. Let $N$ be the ambient manifold for the distribution $\mathcal{D}$. Then for each $p \in N$, there is a cubical (its image in $\mathbb{R}^{n}$ is an open cube) coordinate chart $(U,(x^{1},\dots,x^{n}))$ containing $p$, such that for all $q \in U$, $\mathcal{D}_{q} = span\{ \partial_{1}|_{q}, \dots, \partial_{k}|_{q} \}$. The slices $x^{k+1} = c^{k+1}$, $\dots$, $x^{n} = c^{n}$ then form embedded integral submanifolds of $\mathcal{D}$. These charts are called flat for $\mathcal{D}$.
2) Local structure of integral submanifolds: For every integral submanifold $M$ of $N$, and any chart flat for $\mathcal{D}$, the intersection $U \cap M$ is a countable disjoint union of open subsets of parallel $k$-dimensional slices of $U$, each of which is open in $M$ (in the "inner" topology of $M$) and embedded in $N$.
Note that I identify immersed submanifold $M$ with its image in $N$ (to save letters).
Now to the actual proof of the statement. Let $M$ be any integral submanifold, and let $M \subseteq M'$ for some connected integral submanifold of the distribution $\mathcal{D}$. We will now prove that $M$ is both closed and open in $M'$:
i.) $M$ is closed in $M'$:
By assumption $M$ is closed in $N$, hence it is closed in the subspace topology of $M'$ and consequently also in the "inner" topology of $M'$. We conclude that $M$ is a closed subset of $M'$.
ii.) $M$ is open in $M'$:
First, find a flat chart $(U,(x^{1},\dots,x^{n}))$ for $\mathcal{D}$ containing the point $p$. By 2) above, this means that both $M \cap U$ and $M' \cap U$ are disjoint union of some open subsets (open in the subspace topology of the slices) of parallel $k$-dimensional slices.
Let $S$ be the slice of $U$ containing $p$. As $M \subseteq M'$, from 2) we get the two open subsets $W,W' \subseteq S$, such that $M \cap S = W$ and $M' \cap S = W'$. As $W$ is open in $S$, it is also open in $W'$. But by 2) $W'$ is open in $M'$. This implies that $W$ is also open in $M'$. This statement is in fact quite subtle, using the fact that $W' \subseteq S$ with subspace topology is homeomorphic to the connected component of the open submanifold $M' \cap U \subseteq M'$. Note that by construction $W \subseteq M$.
To summarize, to every $p \in M$, we have found its neighborhood $W$ open in $M'$, such that $W \subseteq M.$ This proves that $M$ is open in $M'$.
As $M'$ is connected, we see that $M = M'$. If $M$ is assumed connected, we have just proved that it is a maximal connected integral submanifold. Q.E.D.
Observe that I have in fact proved a slightly stronger statement - any integral submanifold which is closed as a subset of $N$ and it is contained in some connected integral submanifold $M' \subseteq N$ must be necessarily the whole $M$'.
