For future students, below you can find a more general case, basically the one presented by @Davide Giraudo, but maybe a little more detailed. The proof is taken from http://www.polishedproofs.com/bxy-is-a-banach-space-whenever-y-is-a-banach-space/
Let $X$ and $Y$ be normed linear spaces, and let $B(X,Y)$ denote the collection of all bounded linear operators from $X$ to $Y$ endowed with the operator norm. Show that $B(X,Y)$ is a normed linear space, and $B(X,Y)$ is a Banach space whenever $Y$ is a Banach space. The vector operations in $B(X,Y)$ are defined pointwise, i.e. $(A+B)(x)=Ax+Bx$, and $(\alpha A)(x)=\alpha (Ax)$. (Notice that in your case $X'=B(X,\mathbb{C})$ and $\mathbb{C}$ is a Banach space)
It is clear that linear operators form a linear space. To show that $B(X,Y)$ is a linear subspace, it is enough to show the closure to addition and scalar multiplication. But these follow easily from the properties of a norm (the fact that the operator norm satisfies all the properties of a norm for bounded functionals is an easy exercise that follows from properties of supremums in $[0, \infty)$) , namely for any $A,B \in B(X,Y)$ and $\lambda \in \mathbb{C}$ $\|A+B\| \leq \|A\|+\|B\| < \infty$ $\|\lambda A\|=|\lambda| \cdot \|A\| < \infty$ Thus, $B(X,Y)$ is a normed linear space.
Now assume that $Y$ is a Banach space. Let $\{A_i\}$ be a Cauchy sequence in $B(X,Y)$, i.e. $\forall \, \epsilon >0$, $\exists \, N \in \mathbb{N}$ such that $\forall \, m,n > N$, $\|A_n-A_m\|< \epsilon $. Let $x \in X$ be arbitrary. Let $\epsilon>0$ be arbitrary. If $x=0$, then $\|A_nx-A_mx\|=0<\epsilon.$ If $x \neq 0$, choose $N$ such that $\|A_n-A_m\|< \frac{\epsilon}{\|x\|}$. Then by a property of the operator norm, $\forall \, m,n > N$, \begin{equation} \begin{split} \|A_nx-A_mx\| & = \|(A_n-A_m)x\|\\ & \leq \|(A_n-A_m)\| \cdot \|x\|\\ & < \frac{\epsilon}{\|x\|} \cdot \|x\|\\ & = \epsilon\\ \end{split} \end{equation}
Thus, in both cases $\{A_nx\}$ is a Cauchy sequence in $Y$. Since $Y$ is a Banach space, it is convergent to some element in $Y$. Call that element $Ax$, i.e. $\lim_{n \rightarrow \infty} A_nx=Ax$ Since $x$ was arbitrary, $Ax$ is defined for any $x \in X$. Thus, $A$ is a map from $X$ to $Y$ defined by $x \rightarrow Ax$. We need to show that $A$ is linear, bounded, and $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Notice that $A$ is linear, since by linearity of $A_n$ we get that for any $x_1, x_2 \in X$, $\lambda \in \mathbb{C}$, \begin{equation} \begin{split} A(x_1+x_2) & = \lim_{n \rightarrow \infty} A_n(x_1+x_2)\\ & = \lim_{n \rightarrow \infty} (A_nx_1+A_nx_2)\\ & = \lim_{n \rightarrow \infty} A_nx_1+\lim_{n \rightarrow \infty} A_nx_2\\ & = Ax_1+Ax_2\\ \end{split} \end{equation} \begin{equation} \begin{split} A(\lambda x_1) & = \lim_{n \rightarrow \infty} A_n(\lambda x_1)\\ & = \lim_{n \rightarrow \infty} \lambda \cdot A_nx_1\\ & = \lambda \lim_{n \rightarrow \infty} A_nx_1\\ & = \lambda\cdot Ax_1\\ \end{split} \end{equation}
Now recall that Cauchy sequences are bounded. Thus, $\forall \, n$, $\|A_n\| for some $C \in \mathbb{R}$. Using this fact, we can see that $A$ is bounded, since by continuity of a norm: \begin{equation} \begin{split} \|A\| & =\sup_{\|x\| \leq 1} \|Ax\|\\ & =\sup_{\|x\| \leq 1} \|\lim_{n \rightarrow \infty} A_nx\|\\ & =\sup_{\|x\| \leq 1} \lim_{n \rightarrow \infty} \|A_nx\|\\ & =\sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \|A_nx\|\\ & \leq \sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \Big(\|A_n\|\cdot \|x\|\Big)\\ & \leq \sup_{\|x\| \leq 1} C \cdot \|x\|\\ & = C \sup_{\|x\| \leq 1} \|x\|\\ & \leq C \\ \end{split} \end{equation}
Finally, we want to show that $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Let $\epsilon > 0$ be arbitrary. Recall that for an arbitrary $x \in X$, we have $\|A_nx-A_mx\| \leq \|(A_n-A_m)\| \cdot \|x\|$ Since $\{A_n\}$ is Cauchy, choose $N$ big enough such that for all $n,m \geq N$, $\|(A_n-A_m)\| < \epsilon$. Then the above inequality turns into $\|A_nx-A_mx\| \leq \epsilon \cdot \|x\|$ Now by continuity of a norm, we can take limit on both sides as $m$ goes to infinity to obtain $\|A_nx-Ax\| \leq \epsilon \cdot \|x\|$ Now taking supremum on both sides over all $x$ such that $\|x\| \leq 1$ yields $\sup_{\|x\| \leq 1}\|A_nx-Ax\| \leq \epsilon$ But this is equivalent to saying that for all $n \geq N$, $\|A_n-A\| \leq \epsilon$ And since $\epsilon$ was arbitrary, this implies that $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Thus, we conclude that $B(X,Y)$ is a Banach space.