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If $T$ is diagonalisable. $W$ is a $T$-invariant subspace of $V$ and $dim V=n$, $dim W=k<n$. I have to prove $T|_{W}$ is diagonalisable.

Taking $\mathfrak B'$ is a basis of $W$ then extending it to a basis $\mathfrak B$ of $V$, I have $$[T]_{\mathfrak B}= \begin{pmatrix} B_1 & B_2 \\ 0 & B_3 \\ \end{pmatrix} $$, where $[T]_{\mathfrak B'}=B_1$. If $T$ is diagonalisable $\exists P \in GL(n,F)$ s.t $P[T]_{\mathfrak B}P^{-1}=D_n$ , where $D_n$ is diagonal matrix but this $P$ can't be restricted in sub-parts of $B_1$ only. So, how to prove this?

**I don't want to use this

Existence of T-invariant complement of T-invariant subspace when T is diagonalisable Because I feel this can be proved without using the result.**

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If you want to do it entirely by hand, write (as you did) $T=\begin{pmatrix} T_1 & T_2 \\ 0 & T_3 \end{pmatrix}$. The eigenvectors of $T$ are the vectors $v=(v_1,v_2)$ with $$ T_1 v_1 + T_2 v_2 = \lambda v_1, \quad T_3 v_2 =\lambda v_2 . $$ We obtain two types of eigenvectors: (1) $v_2$ is an eigenvector of $T_3$, and then $v_1=-(T_1-\lambda)^{-1}T_2v_2$. Here I can assume that $T_1,T_3$ don't have eigenvalues in common, so the inverse exists. I slightly perturb $T_3$ if necessary, such that $T$ stays diagonalizable; (2) $v_2=0$, and then $v_1$ must be an eigenvector of $T_1$

I have at most $n-k$ linearly independent eigenvectors of type (1), so since $T$ is diagonalizable, there must be $\ge k$ linearly independent eigenvectors of type (2). This says that $T_1$ is diagonalizable, as desired.