If $V = \bigoplus_{i<n} U_i$ with every $U_i$ finite-dimensional, then $V$ is finite-dimensional and $\dim_F V = \sum_{i<n} \dim_F U_i$; in particular $\dim_F(U \oplus W) = \dim_F U + \dim_F W$

The statement to be proved by induction on $n$ is: for every vector space $V$ over $F$ and every family of finite-dimensional linear subspaces $U_i$ of $V$ indexed by $i < n$ with $V = \bigoplus_{i<n}U_i$, the space $V$ is finite-dimensional with $\dim_F V = \sum_{i<n}\dim_F U_i$. At $n = 0$ the hypothesis $V = \bigoplus_{i<0}U_i$ holds exactly when $V = \{0_V\}$, and then $\dim_F V = 0$, which is also the empty sum $\sum_{i<0}\dim_F U_i$.

Two identities used in the successor step. Let $n = \sigma(p)$ and put $V' := \sum_{i<p}U_i$, a linear subspace of $V$. First, $V' = \sum_{i \ne p}U_i$: an element of the right-hand side is $\sum_{i<\sigma(p)}u_i$ with $u_i \in U_i$ for $i < p$ and $u_p = 0_V$, and the recursion makes that $\bigl(\sum_{i<p}u_i\bigr) + 0_V = \sum_{i<p}u_i$, so the two sets have the same elements. Second, $V' + U_p = \sum_{i<\sigma(p)}U_i$: an element of the left-hand side is $\bigl(\sum_{i<p}u_i\bigr) + u_p$ with $u_i \in U_i$ for $i < p$ and $u_p \in U_p$, which by the recursion is $\sum_{i<\sigma(p)}u_i$, and conversely.

Assume the displayed statement at the natural number $p$, for every vector space over $F$ and every family of finite-dimensional linear subspaces indexed by $i < p$.

The successor step. Let $n = \sigma(p)$, let $V = \bigoplus_{i<\sigma(p)}U_i$ with every $U_i$ finite-dimensional, and let $V' := \sum_{i<p}U_i$ as in step 1.2. Then $V' = \bigoplus_{i<p}U_i$ as a direct sum inside $V'$: condition (D1) holds by the definition of $V'$, and for $j < p$ every element $\sum_{i<p}u_i$ with $u_i \in U^{(j)}_i$ is also $\sum_{i<\sigma(p)}u_i$ after setting $u_p := 0_V \in U_p$, so $\sum_{i<p}U^{(j)}_i$ is contained in the corresponding sum for the family indexed by $\sigma(p)$, and (D2) for the larger family at $j$ forces $U_j \cap \sum_{i<p}U^{(j)}_i = \{0_V\}$, the reverse inclusion holding because both sides are linear subspaces. Each $U_i$ with $i < p$ is contained in $V'$ and is therefore a linear subspace of $V'$, still finite-dimensional. So step 1.3 applies to $V'$ and gives that $V'$ is finite-dimensional with $\dim_F V' = \sum_{i<p}\dim_F U_i$. Now $V'$ and $U_p$ are finite-dimensional linear subspaces of $V$; by step 1.2 their sum is $\sum_{i<\sigma(p)}U_i = V$, and their intersection is $U_p \cap \sum_{i \ne p}U_i = \{0_V\}$ by (D2) at $j = p$. The dimension formula therefore gives $\dim_F V + \dim_F\{0_V\} = \dim_F V' + \dim_F U_p$, that is $\dim_F V = \bigl(\sum_{i<p}\dim_F U_i\bigr) + \dim_F U_p = \sum_{i<\sigma(p)}\dim_F U_i$, and $V$ is finite-dimensional because the dimension formula asserts that the sum of two finite-dimensional subspaces is one.

Step 1.1 and step 2.1 are the base case and the successor step of an induction on $n$, so the statement holds for every $n \in \mathbb{N}$; at $n = 2$ it reads $\dim_F(U_0 \oplus U_1) = \sum_{i<2}\dim_F U_i = \dim_F U_0 + \dim_F U_1$, by the recursion for finite sums of naturals.