Let $V$ be a vector space over $\mathbb{F}$.
Let $T\in\mathcal{L}(V)$.
Suppose that $T$ has a matrix $B=\begin{pmatrix}B_1&&&\\&B_2&&\\&&\ddots&\\&&&B_r\end{pmatrix}$ with respect to some basis $v^{1}_1\dots,v^{1}_{n_1},\dots,v^{r}_1\dots,v^{r}_{n_r}$ of $V$, where $B_i$ is a $n_i\times n_i$ matrix for $i\in\{1,\dots,r\}$.
Let $V_1:=\operatorname{span}(v^{1}_1\dots,v^{1}_{n_1})$.
Let $V_2:=\operatorname{span}(v^{2}_1\dots,v^{2}_{n_2})$.
$\dots$
Let $V_r:=\operatorname{span}(v^{r}_1\dots,v^{r}_{n_r})$.
Prove that $T$ is diagonalizable $\Leftrightarrow$ $T|_{V_1},T|_{V_2},\dots,T|_{V_r}$ are diagonalizable.
I can prove that $T$ is diagonalizable $\Leftarrow$ $T|_{V_1},T|_{V_2},\dots,T|_{V_r}$ are diagonalizable.
Proof:
Suppose that $T|_{V_1},T|_{V_2},\dots,T|_{V_r}$ are diagonalizable.
Then, $T|_{V_i}$ has a diagonal matrix $C_i$ with respect to some basis $w^{i}_1\dots,w^{i}_{n_i}$ of $V_i$ for $i\in\{1,\dots,r\}$.
Then, $C=\begin{pmatrix}C_1&&&\\&C_2&&\\&&\ddots&\\&&&C_r\end{pmatrix}$ is the matrix with respect to the basis $w^{1}_1\dots,w^{1}_{n_1},\dots,w^{r}_1\dots,w^{r}_{n_r}$ of $V$.
$C$ is obviously diagonal.
I cannot prove that $T$ is diagonalizable $\Rightarrow$ $T|_{V_1},T|_{V_2},\dots,T|_{V_r}$ are diagonalizable.
$V_1,V_2,\dots,V_r$ are subspaces of $V$ which are invariant under $T$.
$V=V_1\oplus\dots\oplus V_r$.
Let $\lambda_1,\dots,\lambda_s$ be the distinct eigenvalues of $T$.
Note that the set of the eigenvalues of $T|_{V_j}$ is a subset of $\{\lambda_1,\dots,\lambda_s\}$ for $j\in\{1,\dots,r\}$.
Let $W(\lambda_i):=\{v\in V\mid Tv=\lambda_i v\}$ for $i\in\{1,\dots,s\}$.
Let $W^{j}(\lambda_i):=\{v\in V_j\mid T|_{V_j}v=\lambda_i v\}$ for $i\in\{1,\dots,s\}$ and for $j\in\{1,\dots,r\}$.
Note that it is possible that $W^{j}(\lambda_i)=\{0\}$.
Since $W^1(\lambda_i)\subset V_1,\dots,W^r(\lambda_i)\subset V_r$ and $V_1+\dots+V_r$ is a direct sum, $W^1(\lambda_i)+\dots+ W^r(\lambda_i)$ is a direct sum.
Then, $W(\lambda_i)=W^1(\lambda_i)\oplus\dots\oplus W^r(\lambda_i)$ for $i\in\{1,\dots,s\}$.
Proof:
Let $v\in W(\lambda_i)$.
Then, $Tv=\lambda_i v$.
We can write $v=v_1+\dots+v_r$, where $v_j\in V_j$ for $j\in\{1,\dots,r\}$.
$Tv=Tv_1+\dots+Tv_r$ and $Tv_j\in V_j$ for $j\in\{1,\dots,r\}$.
$\lambda_i v=\lambda_i v_1+\dots+\lambda_i v_r$ and $\lambda_i v_j\in V_j$ for $j\in\{1,\dots,r\}$.
Since $V=V_1\oplus\dots\oplus V_r$, $Tv_1=T|_{V_1} v_1=\lambda_i v_1,\dots Tv_r=T|_{V_r} v_r=\lambda_i v_r$.
So, $v_1\in W^1(\lambda_i),\dots,v_r\in W^r(\lambda_i)$.
So, $v=v_1+\dots+v_r\in W^1(\lambda_i)\oplus\dots\oplus W^r(\lambda_i)$.
Conversely, let $v\in W^1(\lambda_i)\oplus\dots\oplus W^r(\lambda_i)$.
We can write $v=v_1+\dots+v_r$, where $v_j\in W^j(\lambda_i)$ for $j\in\{1,\dots,r\}$.
$Tv=Tv_1+\dots+Tv_r=T|_{V_1} v_1+\dots+T|_{V_r} v_r=\lambda_i v_1+\dots+\lambda_i v_r=\lambda_i(v_1+\dots+v_r)=\lambda_i v$.
So, $v\in W(\lambda_i)$.
So, $\dim W(\lambda_i)=\dim W^1(\lambda_i)+\dots+ \dim W^r(\lambda_i)$ for $i\in\{1,\dots,s\}$.
$T$ is diagonalizable.
$\Leftrightarrow$
$\dim V=\dim V_1+\dots+\dim V_r=\dim W(\lambda_1)+\dots+\dim W(\lambda_s)$.
$\Leftrightarrow$
$\dim V=\dim V_1+\dots+\dim V_r=\sum_{j=1}^{r}\dim W^j(\lambda_1)+\dots+\sum_{j=1}^{r}\dim W^j(\lambda_s).$
$\Leftrightarrow$
$\dim V=\dim V_1+\dots+\dim V_r=(\dim W^1(\lambda_1)+\dots+\dim W^1(\lambda_s))+\dots+(\dim W^r(\lambda_1)+\dots+\dim W^r(\lambda_s)).$
$\Leftrightarrow$
$\dim V_1=\dim W^1(\lambda_1)+\dots+\dim W^1(\lambda_s),\dots,\dim V_r=\dim W^r(\lambda_1)+\dots+\dim W^r(\lambda_s).$
$\Leftrightarrow$
$T|_{V_1},\dots,T|_{V_r}$ are diagonalizable.