The usual lower bound (e.g., Theorem II.13 in Davidson and Szarek ("Banach space theory and local operator theory") says that if $A\in R^{n\times d}$ has iid $N(0,1)$ entries then $$ P(\sigma_{\min}(A^TA)^{1/2} \ge \sqrt n - \sqrt d - t) \ge 1- e^{-t^2/2}. $$

You may assume $\sigma_{\min}(\Sigma)>0$ (that is, $\Sigma$ invertible), otherwise there is nothing prove. Now let $A=X\Sigma^{-1/2}$ so that $A$ has iid $N(0,1)$ entries as above. Then $$\sigma_{\min}(X^TX)=\sigma_{\min}(\Sigma^{1/2}A^TA\Sigma^{1/2}) \ge \sigma_{\min}(\Sigma) \sigma_{\min}(A^TA). $$ Applying the concentration inequality for $\sigma_{\min}(A^TA)$ gives the desired bound.

The usual lower bound (e.g., Theorem II.13 in Davidson and Szarek ("Banach space theory and local operator theory") says that if $A\in R^{n\times d}$ has iid $N(0,1)$ entries then $$ P(\sigma_{\min}(A^TA)^{1/2} \ge \sqrt n - \sqrt d - t) \ge 1- e^{-t^2/2}. $$

You may assume $\sigma_{\min}(\Sigma)>0$ (that is, $\Sigma$ invertible), otherwise there is nothing prove. Now let $A=X\Sigma^{-1/2}$ so that $A$ has iid $N(0,1)$ entries as above. Then $$\sigma_{\min}(X^TX)=\sigma_{\min}(\Sigma^{1/2}A^TA\Sigma^{1/2}) \ge \sigma_{\min}(\Sigma) \sigma_{\min}(A^TA). $$ This is because, if the vector $u$ with $\|u\|=1$ is such that $\|Au\|^2=\sigma_{\min}(A^TA)$, then $\|Au\|^2=\|X\Sigma^{-1/2}u\|^2\le\sigma_{\min}(X^TX)\|\Sigma^{-1/2}u\|^2$ and $\|\Sigma^{-1/2}u\|^2\le \|\Sigma^{-1}\|_{operator}=\sigma_{\min}(\Sigma)^{-1}$.

Applying the concentration inequality for $\sigma_{\min}(A^TA)$ gives the desired bound.