Recall that the sum of subspaces $U$ and $V$ is
\[U+V=\{\mathbf{x}+\mathbf{y} \mid \mathbf{x}\in U, \mathbf{y}\in V\}.\]
The sum $U+V$ is a subspace.
(See the post “The sum of subspaces is a subspace of a vector space” for a proof.)

Proof.

Let $n=\dim(U)$ and $m=\dim(V)$.
Let
\[B_1=\{\mathbf{u}_1, \dots, \mathbf{u}_n\}\]
be a basis of the vector space $U$ and let
\[B_2=\{\mathbf{v}_1, \dots, \mathbf{v}_m\}\]
be a basis of the vector space $V$.

An arbitrary element of the vector space $U+W$ is of the form $\mathbf{x}+\mathbf{y}$, where $\mathbf{x}\in U$ and $\mathbf{y} \in V$.

Since $B_1$ is a basis of $U$, we can write
\[\mathbf{x}=r_1\mathbf{u}_1+\cdots +r_n \mathbf{u}_n\]
for some scalars $r_1, \dots, r_n\in K$.
Also, since $B_2$ is a basis of $V$, we can write
\[\mathbf{y}=s_1\mathbf{v}_1+\cdots +s_m \mathbf{v}_m\]
for some scalars $s_1, \dots, s_m\in K$.

Thus, we have
\begin{align*}
\mathbf{x}+\mathbf{y}&=r_1\mathbf{u}_1+\cdots +r_n \mathbf{u}_n+s_1\mathbf{v}_1+\cdots +s_m \mathbf{v}_m,
\end{align*}
and hence $\mathbf{x}+\mathbf{y}$ is in the span $S:=\Span(\mathbf{u}_1, \dots, \mathbf{u}_n, \mathbf{v}_1, \dots, \mathbf{v}_m)$.

Thus we have $U+W \subset S$ and it yields that \begin{align*}
\dim(U+W) \leq \dim(S)\leq n+m=\dim(U)+\dim(V).
\end{align*}
This completes the proof.

Related Question.

Let $A$ and $B$ be $m\times n$ matrices.
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\[\rk(A+B) \leq \rk(A)+\rk(B).\]

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[…] Here we used the fact that for two subspaces $U$ and $V$ of a vector space, we have [dim(U+V)leq dim(U)+dim(V).] For a proof of this fact, see the post “Dimension of the sum of two subspaces“. […]

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[…] For a proof, see the post “Dimension of the sum of two subspaces“. […]

[…] Here we used the fact that for two subspaces $U$ and $V$ of a vector space, we have [dim(U+V)leq dim(U)+dim(V).] For a proof of this fact, see the post “Dimension of the sum of two subspaces“. […]