A proof can be found in this article by Hildebrandt, Jost, and Widman. I reproduce here the proof for completeness.
Consider the usual representation of $\mathfrak{m}$ as matrices of the form
$$ \mathfrak{m} = \left\{ \begin{pmatrix} 0 & A \\ -A^\intercal & 0 \end{pmatrix} \bigm\vert A \in \mathbb{R}^{(n-k)\times k} \right\} $$
We can write the sectional curvature of the Grassmannian at $A, B \in \mathbb{R}^{(n-k)\times k}$ for two matrices such that $\operatorname{tr}(A^\intercal B)$ in terms of their Frobenius norm as
$$ \operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) = \frac{ \lVert AB^\intercal - BA^\intercal \rVert_F^2 + \lVert A^\intercal B - B^\intercal A \rVert_F^2 }{ 2\lVert A \rVert_F^2 \lVert B \rVert_F^2 } $$
Considering the SVD of A, we may assume that $A$ just has non-zero elements in its main diagonal. We can then write the numerator of the sectional curvature as
$$ N = \sum_{i\neq j} (a_{ii}b_{ji}-a_{jj}b_{ij})^2 + \sum_{i\neq j} (a_{ii}b_{ij}-a_{jj}b_{ji})^2 $$ $$ D = 2(\sum_i a_{ii}^2)(\sum_{j,k}b_{jk}^2) $$
We can bound $N$ using $(a+b)^2 \leq 2(a^2 + b^2)$ on the summation terms and Cauchy-Schwarz so that
$$ N \leq 4\sum_{i\neq j} a_{ii}^2b_{ji}^2 + 4\sum_{i\neq j} a_{ii}^2b_{ij}^2 \leq 4(\sum_{i} a_{ii}^2)(\sum_{j \neq i}b_{ji}^2 + \sum_{j \neq i}b_{ij}^2) $$
and we can bound the denominator as
$$ D = 2(\sum_i a_{ii}^2)(\sum_j b^2_{ji} + \sum_k\sum_{j \neq i}b_{kj}^2) \geq 2(\sum_{i} a_{ii}^2)(\sum_j b^2_{ji} + \sum_k\sum_{j \neq i}b_{kj}^2) \geq \frac{1}{2}N. $$
In the paper they also show the tightness of this bound considering $A = \mathrm{Id}$ and $B =\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$.