If $T_1, T_2$ are linear compact operators, then $T_1+T_2$ is also linear and compact. Let $X,Y$ be Banach space.
Prove that if $T_1 : X\to Y, T_2 : X\to Y$ are both linear and compact operators, then $T_1+T_2$ is also linear and compact.
The definition of compact operator is here.

Compact Operator
$T:X\to Y$ is compact iff for every bounded sequence $\{x_n\}_{n=1}^\infty\subset X$, there exists a subsequence of $\{Tx_n\}_{n=1}^\infty$ converging to an element in $Y.$

Let me prove.
The linearity of $T_1+T_2$ is almost obvious from the linearity of $T_1,T_2$ and the definition of sum of operator, so I should prove the compactness of $T_1+T_2.$
Let $\{x_n\}_{n=1}^\infty\subset X$ be bounded.
Since $T_1$ is compact, there exist
$\{T_1 y_n\}_{n=1}^\infty\  $ : subsequence of $\{ T_1 x_n \}_{n=1}^\infty\ \ $ and $\lambda \in Y$
such that $T_1 y_n\to \lambda$ as $n\to \infty.$
Now, $\{y_n \}$ is a subsequence of $\{x_n\}$ and since $\{x_n\}$ is bounded, $\{y_n\}$ is also bounded.
From the compactness of $T_2$, there exist
$\{T_2 z_n\}_{n=1}^\infty\  $ : subsequence of $\{ T_2 y_n \}_{n=1}^\infty\ \ $ and $\mu \in Y$
such that $T_2 z_n \to \mu$ as $n\to \infty$.
Then, $(T_1+T_2)z_n=T_1 z_n +T_2z_n \to \lambda +\mu\in Y,$ where $T_1 z_n\to \lambda$ is from the fact that $\{T_1 z_n\}$ is a subsequence of $\{T_1 y_n\}$ and $T_1 y_n\to \lambda,$ and $\lambda+\mu\in Y$ is from the fact that $Y$ is vector space.
This shows that $\{ (T_1+T_2)z_n\}_{n=1}^\infty$ is a subsequence of $\{ (T_1+T_2)x_n\}_{n=1}^\infty$ converging to an element of $Y,$ thus $T_1+T_2$ is compact.

Does this proof make sense ?
Please point out if there are any mistakes and/or errors.

Alt:
$T\in \mathcal{L}(X, Y) $ is compact iff $T(\overline{B(0, 1) }) \subset Y$ is compact.
$(T_1+T_2)(\overline{B(0, 1) })=T_1(\overline{B(0, 1) })+T_2(\overline{B(0, 1) })$
Now it follows from Compactnesss of $T_1$ and $T_2$ that $T_1+T_2$ is compact.
