If $V={⨁}_{i<n}{U}_i$ 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 $V={⨁}_{i<n}{U}_i$, the space $V$ is finite-dimensional with ${\dim }_{F}V={\sum }_{i<n}{\dim }_F{U}_i$. At $n=0$ the hypothesis $V={⨁}_{i<0}{U}_i$ 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 $V^{\mathrm{\prime }}:={\sum }_{i<p}{U}_i$, a linear subspace of $V$. First, $V^{\mathrm{\prime }}={\sum }_{i\mathrm{\ne }p}{U}_i$: 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 $\left({\sum }_{i<p}{u}_i\right)+0_V={\sum }_{i<p}{u}_i$, so the two sets have the same elements. Second, $V^{\mathrm{\prime }}+U_p={\sum }_{i<\sigma (p)}{U}_i$: an element of the left-hand side is $\left({\sum }_{i<p}{u}_i\right)+u_p$ 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 $V={⨁}_{i<\sigma (p)}{U}_i$ with every $U_i$ finite-dimensional, and let $V^{\mathrm{\prime }}:={\sum }_{i<p}{U}_i$ as in step 1.2. Then $V^{\mathrm{\prime }}={⨁}_{i<p}{U}_i$ as a direct sum inside $V^{\mathrm{\prime }}$: condition (D1) holds by the definition of $V^{\mathrm{\prime }}$, and for $j<p$ every element ${\sum }_{i<p}{u}_i$ with $u_i\in U_i^{(j)}$ is also ${\sum }_{i<\sigma (p)}{u}_i$ after setting $u_p:=0_V\in U_p$, so ${\sum }_{i<p}{U}_i^{(j)}$ is contained in the corresponding sum for the family indexed by $\sigma (p)$, and (D2) for the larger family at $j$ forces $U_j\cap {\sum }_{i<p}{U}_i^{(j)}=\{0_V\}$, the reverse inclusion holding because both sides are linear subspaces. Each $U_i$ with $i<p$ is contained in $V^{\mathrm{\prime }}$ and is therefore a linear subspace of $V^{\mathrm{\prime }}$, still finite-dimensional. So step 1.3 applies to $V^{\mathrm{\prime }}$ and gives that $V^{\mathrm{\prime }}$ is finite-dimensional with ${\dim }_F{V}^{\mathrm{\prime }}={\sum }_{i<p}{\dim }_F{U}_i$. Now $V^{\mathrm{\prime }}$ 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 $U_p\cap {\sum }_{i\mathrm{\ne }p}{U}_i=\{0_V\}$ by (D2) at $j=p$. The dimension formula therefore gives ${\dim }_{F}V+{\dim }_F\{0_V\}={\dim }_F{V}^{\mathrm{\prime }}+{\dim }_F{U}_p$, that is ${\dim }_{F}V=\left({\sum }_{i<p}{\dim }_F{U}_i\right)+{\dim }_F{U}_p={\sum }_{i<\sigma (p)}{\dim }_F{U}_i$, 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.