A proof can be found in [this article by Hildebrandt, Jost, and Widman][1]. 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

$$
\operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) = 
\frac{
\lVert AB^\intercal - BA^\intercal \rVert_F^2 + 
\lVert A^\intercal B - B^\intercal A \rVert_F^2
}{
2\lVert A \rVert_F^2 \lVert B \rVert_F^2
}
$$

In the rest, we will be deliberately imprecise with the limits of the indices to not clutter innecessarily the proof.

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 $B =\begin{pmatrix}
    0 & -1 \\
    1 & 0
\end{pmatrix}$.


  [1]: https://link.springer.com/content/pdf/10.1007/BF01389161.pdf