Iosif Pinelis neatly answered the question. I just want to add another viewpoint. The Minkowski functional (or gauge) $p_B(x)=\inf\{t>0: x\in tB\}$ of an absolutely convex absorbing set is a semi-norm which almost gives back $B$ since $\{p_B<1\}\subseteq B\subseteq \{p_B\le 1\}$. 

Since two semi-norms $p_k$ on $V$ with unit balls $A_k=\{p_k\le 1\}$ satisfy $p_1\le p_2$ if and only if $A_2\subseteq A_1$, it is immediate that the Minkowski functional of the convex (=absolutely convex) hull of $B_1\cup B_2$ (where $B_k$ are the unit balls of $\|\cdot\|_k$) is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$ in the partially ordered set of semi-norms on $V$. One has thus to check that $\|\cdot\|_+$ is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$: On the one hand, $\|v\|_+\le\|v\|_k$ since $v=0+v=v+0$ and, on the other hand, if $p$ is a seminorm with $p\le\|\cdot\|_k$ the triangle inequality yields $p\le\|\cdot\|_+$.


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I doubt, by the way, that $\|\cdot\|_+$ is always a proper norm
(it is of course a semi-norm): Any two separable Banach spaces have the same algebraic dimension (namely, the cardinality of $\mathbb R$, see, e.g.,
https://math.stackexchange.com/questions/1899365/hamel-dimension-of-infinite-dimensional-separable-banach-space,
however, for a simple example, consider $\ell^2\subseteq c_0$, so that dim$(\ell^2)\le$dim$(c_0)$, but on the other hand, $c_0\to\ell^2$, $(x_n)_{n\in\mathbb N}\mapsto (x_n/n)_{n\in\mathbb N}$ is a linear injection which implies dim$(c_0)\le$dim$(\ell^2)$, talking about Hamel dimension requires the axiom of choice). We can thus find two incomparable complete norms $\|\cdot\|_k$ on a vector space $V$. If $\|\cdot\|_+$ were a proper norm, the identical map $i:(V,\|\cdot\|_1)\to (V,\|\cdot\|_2)$ would have closed graph because it is continuous if both spaces are endowed with the coarser Hausdorff topology generated by $\|\cdot\|_+$ (if one prefers an elementary argument, for $x_n\to x$ in $(V;\|\cdot\|_1)$ with $i(x_n)\to y$ in $(V,\|\cdot\|_2)$ we have $i(x)=y$ because $\|\cdot\|_k$-convergence implies $\|\cdot\|_+$-convergence and $\|\cdot\|_+$-limits are unique). The closed graph theorem implies that $i$ is continuous so that $\|\cdot\|_2\le c\|\cdot\|_1$ for some constant $c$, contradicting the incomparability of the norms.

When talking about *closed* unit balls one should thus state precisely the topology in which the ball is closed.