The following construction shows that such a quadruple does indeed exist. Some of the ideas in this proof are inspired by the content from Section 1.15 of Introduction to Hilbert Space and the Theory of Spectral Multiplicity (Second Edition) by Halmos. Also see here.
Let $\mathcal{H}$ be any separable infinite-dimensional Hilbert space. Let $(e_{n})_{n\in\mathbb{N}}$ and $(f_{n})_{n\in\mathbb{N}}$ be orthonormal sequences in $\mathcal{H}$ such that $(e_{n}, f_{m} )_{\mathcal{H}} = 0$ for all $m,n\in\mathbb{N}$ and $\overline{{\rm span}}(\{e_{n} : n\in\mathbb{N} \} \cup\{ f_{n} : n\in\mathbb{N}\}) = \mathcal{H}$. Define $g_{n} := (1 - \tfrac{1}{n^{4}})^{\tfrac{1}{2}} e_{n} + \tfrac{1}{n^{2}} f_{n}$ for each $n\in\mathbb{N}$. It is straightforward to check that $(g_{n})_{n\in\mathbb{N}}$ is an orthonormal sequence. Define $u := \sum_{n\in\mathbb{N}} \tfrac{1}{n^{2}}f_{n}$ and $v := \sum_{n\in\mathbb{N}} \tfrac{1}{n} f_{n}$, where both sums converge (unconditionally) in $\mathcal{H}$ since $(\tfrac{1}{n^{2}})_{n\in\mathbb{N}}$ and $(\tfrac{1}{n})_{n\in\mathbb{N}}$ both belong to $\ell^{2}$.
Define $\mathcal{K}_{1} := \overline{{\rm span}}\{e_{n} : n\in\mathbb{N}\}$, $\mathcal{H}_{1} := \mathcal{K}_{1} + \mathbb{K}v$, $\mathcal{K}_{2} := \overline{{\rm span}}\{g_{n} : n\in\mathbb{N}\}$ and $\mathcal{H}_{2} := \mathcal{K}_{2} + \mathbb{K}u$. It is immediate that $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$ are closed subspaces of $\mathcal{H}$, hence so are their finite-dimensional extensions $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$. We claim the following properties hold.
\begin{equation}
\mathcal{K}_{1} \text{ is a proper closed subspace of } \mathcal{H}_{1} , \tag{1}
\end{equation}
\begin{equation}
\mathcal{K}_{2} \text{ is a proper closed subspace of } \mathcal{H}_{2} , \tag{2}
\end{equation}
\begin{equation}
\mathcal{H}_{1}\cap \mathcal{H}_{2} = \{0\}, \text{ and} \tag{3}
\end{equation}
\begin{equation}
\overline{{\rm span}}(\mathcal{K}_{1} \cup \mathcal{K}_{2}) = \overline{{\rm span}}(\mathcal{H}_{1} \cup \mathcal{H}_{2}) = \mathcal{H} . \tag{4}
\end{equation}
We first show $(4)$. With the inclusion $\overline{{\rm span}}(\mathcal{K}_{1} \cup \mathcal{K}_{2}) \subseteq \overline{{\rm span}}(\mathcal{H}_{1} \cup \mathcal{H}_{2})$ in mind it is sufficient to show $\overline{{\rm span}}(\mathcal{K}_{1} \cup \mathcal{K}_{2}) = \mathcal{H}$. By the definition of $\mathcal{K}_{1}$ we have $e_{n} \in \mathcal{K}_{1} \subseteq \overline{{\rm span}}(\mathcal{K}_{1} \cup \mathcal{K}_{2})$ for each $n\in\mathbb{N}$. Moreover, we also have $f_{n} = - n^{2} (1 - \tfrac{1}{n^{4}})^{\tfrac{1}{2}} e_{n} + n^{2} g_{n} \in \mathcal{K}_{1} + \mathcal{K}_{2} \subseteq \overline{{\rm span}}(\mathcal{K}_{1} \cup \mathcal{K}_{2})$ for each $n\in\mathbb{N}$. It follows that $\mathcal{H} = \overline{{\rm span}}(\{e_{n} : n\in\mathbb{N} \} \cup\{ f_{n} : n\in\mathbb{N}\}) \subseteq \overline{{\rm span}}(\mathcal{K}_{1} \cup \mathcal{K}_{2})$. This shows $(4)$.
We next show that $u\not\in \mathcal{K}_{1} + \mathcal{K}_{2}$ and $v\not\in \mathcal{K}_{1} + \mathcal{K}_{2}$. Once this is done, we obtain $v\in \mathcal{H}_{1}\setminus \mathcal{K}_{1}$ and $u\in \mathcal{H}_{2}\setminus \mathcal{K}_{2}$, which show $(1)$ and $(2)$ respectively.
Suppose for a contradiction that $u\in \mathcal{K}_{1} + \mathcal{K}_{2}$. Write $u = x + y$ where $x\in \mathcal{K}_{1}$ and $y\in \mathcal{K}_{2}$. Let $n\in\mathbb{N}$. Observe that $f_{n}\in \mathcal{K}_{1}^{\perp}$, $(g_{n}, f_{n})_{\mathcal{H}} = \tfrac{1}{n^{2}}$ and $(g_{m}, f_{n})_{\mathcal{H}} = 0$ for all $m\in\mathbb{N}\setminus\{n\}$. Hence we have
\begin{align*}
\tfrac{1}{n^{2}} = (u, f_{n})_{\mathcal{H}} = (x + y, f_{n})_{\mathcal{H}} = (y, f_{n})_{\mathcal{H}} &= (\sum_{k\in\mathbb{N}} (y, g_{k})_{\mathcal{H}} g_{k}, f_{n} )_{\mathcal{H}} \\
&= (y, g_{n})_{\mathcal{H}} (g_{n}, f_{n})_{\mathcal{H}} = \tfrac{1}{n^{2}} (y, g_{n})_{\mathcal{H}} .
\end{align*}
This implies $(y, g_{n})_{\mathcal{H}} = 1$ for all $n\in\mathbb{N}$ which contradicts the Bessel inequality. Hence $u\not\in \mathcal{K}_{1} + \mathcal{K}_{2}$.
Similarly, suppose for a contradiction that $v\in \mathcal{K}_{1} + \mathcal{K}_{2}$. Write $v = x + y$ where $x\in \mathcal{K}_{1}$ and $y\in \mathcal{K}_{2}$. For each $n\in\mathbb{N}$ we have
\begin{align*}
\tfrac{1}{n} = (v, f_{n})_{\mathcal{H}} = (x + y, f_{n})_{\mathcal{H}} = (y, f_{n})_{\mathcal{H}} &= (\sum_{k\in\mathbb{N}} (y, g_{k})_{\mathcal{H}} g_{k}, f_{n} )_{\mathcal{H}} \\
&= (y, g_{n})_{\mathcal{H}} (g_{n}, f_{n})_{\mathcal{H}} = \tfrac{1}{n^{2}} (y, g_{n})_{\mathcal{H}} .
\end{align*}
This implies $(y, g_{n})_{\mathcal{H}} = n$ for all $n\in\mathbb{N}$ which contradicts the Bessel inequality. Hence $v\not\in\ \mathcal{K}_{1} + \mathcal{K}_{2}$.
All that is left is to show $(3)$. We do this now. Let $z\in \mathcal{H}_{1}\cap \mathcal{H}_{2}$. Write $z = \alpha v + x = \beta u + y$ with $\alpha , \beta \in \mathbb{K}$, $x\in \mathcal{K}_{1}$ and $y\in \mathcal{K}_{2}$. For each $n\in\mathbb{N}$ we have
\begin{align*}
\tfrac{\alpha}{n} = \alpha (v, f_{n})_{\mathcal{H}} &= (\alpha v + x, f_{n})_{\mathcal{H}} \\
&= (\beta u + y, f_{n})_{\mathcal{H}} \\
&= \beta (u, f_{n})_{\mathcal{H}} + (\sum_{k\in\mathbb{N}} (y, g_{k})_{\mathcal{H}} g_{k}, f_{n} )_{\mathcal{H}} \\
&= \beta (u, f_{n})_{\mathcal{H}} + (y, g_{n})_{\mathcal{H}} (g_{n}, f_{n})_{\mathcal{H}} = \tfrac{1}{n^{2}} (\beta + (y, g_{n})_{\mathcal{H}}) .
\end{align*}
It follows that $(y, g_{n})_{\mathcal{H}} = \alpha n - \beta$ for all $n\in\mathbb{N}$. By the Bessel inequality the sequence $(\alpha n - \beta )_{n\in\mathbb{N}}$ converges to $0$. Hence we must have $\alpha = 0$ and $\beta = 0$. Consequently, we have $(y, g_{n})_{\mathcal{H}} = 0$ for all $n\in\mathbb{N}$ and deduce from the definition of $\mathcal{K}_{2}$ that $y = 0$, so that $z = 0\cdot u + 0 = 0$. Since $0\in \mathcal{H}_{1}\cap \mathcal{H}_{2}$, we deduce $\mathcal{H}_{1} \cap \mathcal{H}_{2} = \{0\}$ and this shows $(4)$.
