How to show that $dX_{t} = \sqrt{2c\lambda} dB_{t}-\lambda X_{t}dt$ has the following solution

If you enter the expression for $dX_s$ into $$ X_0+\int_0^te^{\lambda s}\,dX_s+\int_0^t X_s\lambda e^{\lambda s}\,ds $$ you get \begin{align} e^{\lambda t}X_t&=X_0+\int_0^te^{\lambda s}\sqrt{2c\lambda}\,dB_s-\int_0^te^{\lambda s}\lambda X_s\,ds+\int_0^tX_s\lambda e^{\lambda s}\,ds\\ &=X_0+\int_0^te^{\lambda s}\sqrt{2c\lambda}\,dB_s \end{align} which immediately leads to the solution.

Alternatively, you can apply Ito's formula to the proposed solution \begin{align} X_t&=X_0e^{-\lambda t}+\sqrt{2c\lambda}\int_0^te^{-\lambda(t-s)}\,dB_s\\ &=e^{-\lambda t}\left\{X_0+\sqrt{2c\lambda}\int_0^te^{\lambda s}\,dB_s\right\} \end{align} which gives \begin{align} dX_t&=-\lambda e^{-\lambda t}\left\{X_0+\sqrt{2c\lambda}\int_0^te^{\lambda s}\,dB_s\right\}\,dt+e^{-\lambda t}\sqrt{2c\lambda}e^{\lambda t}\,dB_t\\ &=-\lambda X_t\,dt+\sqrt{2c\lambda}\,dB_t\,. \end{align}