I'm not sure this answers your question since I think that "by hands" might have different meanings. But here is a proof that uses exact sequences.
The Atiyah class of $E$ is the class of the following exact sequence: $$ 0\to E\to Diff^{\leq 1}_X(E)\to T_X\otimes E\to 0 \qquad(1) $$ where $Diff^{\leq 1}_X(E)$ is the sheaf of $E$-valued differential operators of degree $\leq1$.
In the case $E=T_X$ this gives
$$
0\to T_X\to Diff^{\leq 1}_X(T_X)\to T_X\otimes T_X\to 0\qquad(2)
$$
and we have a morphism from it to the following one:
$$
0\to T_X\to Diff^{+,\leq 2}_X\to S^2(T_X)\to 0\qquad(3)
$$
where $Diff^{+,\leq 2}_X$ is the sheaf of differential operators (with values in $\mathcal O_X$) of degree $\leq2$ and vanishing on constants. Moreover one can prove that
$$
T_X=\ker\left(T_X\oplus Diff^{\leq 1}_X(T_X)\to Diff^{+,\leq 2}_X\right)\qquad(4)
$$
Hence the result.
Here is a more geometric interpretation of the above.
Exact sequence (1) splits if and only if there exists a holomorphic connection in $E$.
Exact sequence (3) splits if and only if there exists a torsion free holomorphic connection in $T_X$.
(4) insures that (2) splits if and only if (3) does.